# [SOLVED]Normal Form and Canonical Form of a Quadratic

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi everyone, Take a look at the following question.

Problem:

Determine which of the following quadratic functions $$q_1,\,q_2:\,V\rightarrow\Re$$ is positive definite and find a basis of $$V$$ where one of $$q_1,\,q_2$$ has normal form and the other canonical: $$q_1(x_1,\,x_2)=x_{1}^{2}+2x_1 x_2+2x_{2}^{2}$$ and $$q_{2}(x_1,\,x_2)=x_{1}^{2}-2x_1 x_2-4x_{2}^{2}$$.

My Ideas:

The first part of finding the positive definiteness is easy and by writing these quadratic functions as $$x^{t}Ax$$ and calculating the eigenvalues of $$A$$ we can find out that $$q_{1}$$ is positive definite and $$q_{2}$$ is not.

However then to find a basis where one of these has normal form (having only squared terms with coefficients $$\pm 1$$ or $$0$$) and the other having the canonical form (having only squared terms), in my opinion is impossible.

$$q_1(x_1,\,x_2)=(x_1+x_2)^2+x_{2}^2$$

$q_2(x_1,\,x_2)=(x_1-x_2)^2-5x_{2}^2$

Now if we choose $$y=x_1+x_2$$ and $$z=x_1-x_2$$ then $$q_1$$ transforms to it's normal form but $$q_{2}$$ does not transform to it's canonical form. If $$q_{2}$$ was instead given as $$q_{2}(x_1,\,x_2)=x_{1}^{2}\color{red}{+}2x_1 x_2-4x_{2}^{2}$$ then the above transform works since substituting $$y=x_1+x_2$$ and $$z=x_1-x_2$$ we get,

$q_1(x_1,\,x_2)=(x_1+x_2)^2+x_{2}^2=y^2+z^2$

and

$q_2(x_1,\,x_2)=(x_1+x_2)^2-5x_{2}^2=y^2-5z^2$

So I am assuming that this problem has a typo. Can anybody confirm my assumption or tell me where I went wrong. #### Klaas van Aarsen

##### MHB Seeker
Staff member
Hi everyone, Take a look at the following question.

Problem:

Determine which of the following quadratic functions $$q_1,\,q_2:\,V\rightarrow\Re$$ is positive definite and find a basis of $$V$$ where one of $$q_1,\,q_2$$ has normal form and the other canonical: $$q_1(x_1,\,x_2)=x_{1}^{2}+2x_1 x_2+2x_{2}^{2}$$ and $$q_{2}(x_1,\,x_2)=x_{1}^{2}-2x_1 x_2-4x_{2}^{2}$$.

My Ideas:

The first part of finding the positive definiteness is easy and by writing these quadratic functions as $$x^{t}Ax$$ and calculating the eigenvalues of $$A$$ we can find out that $$q_{1}$$ is positive definite and $$q_{2}$$ is not.

However then to find a basis where one of these has normal form (having only squared terms with coefficients $$\pm 1$$ or $$0$$) and the other having the canonical form (having only squared terms), in my opinion is impossible.

$$q_1(x_1,\,x_2)=(x_1+x_2)^2+x_{2}^2$$

$q_2(x_1,\,x_2)=(x_1-x_2)^2-5x_{2}^2$

Now if we choose $$y=x_1+x_2$$ and $$z=x_1-x_2$$ then $$q_1$$ transforms to it's normal form but $$q_{2}$$ does not transform to it's canonical form. If $$q_{2}$$ was instead given as $$q_{2}(x_1,\,x_2)=x_{1}^{2}\color{red}{+}2x_1 x_2-4x_{2}^{2}$$ then the above transform works since substituting $$y=x_1+x_2$$ and $$z=x_1-x_2$$ we get,

$q_1(x_1,\,x_2)=(x_1+x_2)^2+x_{2}^2=y^2+z^2$

and

$q_2(x_1,\,x_2)=(x_1+x_2)^2-5x_{2}^2=y^2-5z^2$

So I am assuming that this problem has a typo. Can anybody confirm my assumption or tell me where I went wrong. Suppose you have:
$$q(x,y)=\begin{pmatrix} x&y \end{pmatrix} \begin{pmatrix} a^2 & 0 \\ 0 & b^2 \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} =a^2 x^2 + b^2 y^2$$
Then $x'=ax,\ y'=by$ will bring it to normal form.

Suppose you have:
$$q(x,y)=\begin{pmatrix} x&y \end{pmatrix} B D B^T \begin{pmatrix} x\\y \end{pmatrix}$$
where D is a diagonal matrix and B is an orthogonal matrix.
Then:
$$\begin{pmatrix} x'\\y' \end{pmatrix} = B^T \begin{pmatrix} x\\y \end{pmatrix}$$
will bring it one step closer to normal form.

#### Sudharaka

##### Well-known member
MHB Math Helper
Suppose you have:
$$q(x,y)=\begin{pmatrix} x&y \end{pmatrix} \begin{pmatrix} a^2 & 0 \\ 0 & b^2 \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} =a^2 x^2 + b^2 y^2$$
Then $x'=ax,\ y'=by$ will bring it to normal form.

Suppose you have:
$$q(x,y)=\begin{pmatrix} x&y \end{pmatrix} B D B^T \begin{pmatrix} x\\y \end{pmatrix}$$
where D is a diagonal matrix and B is an orthogonal matrix.
Then:
$$\begin{pmatrix} x'\\y' \end{pmatrix} = B^T \begin{pmatrix} x\\y \end{pmatrix}$$
will bring it one step closer to normal form.
Thanks very much for the reply, but the problem is when we bring one to it's normal form the other won't transform itself to the canonical form under the same transformation.If we have two different transformations we can bring one to it's normal form and the other to it's canonical form. Am I wrong? #### Klaas van Aarsen

##### MHB Seeker
Staff member
Thanks very much for the reply, but the problem is when we bring one to it's normal form the other won't transform itself to the canonical form under the same transformation.If we have two different transformations we can bring one to it's normal form and the other to it's canonical form. Am I wrong? I interpret your problem statement as having a different basis for each of the quadratics.
The first one, which is positive definite, can be converted to normal form using the procedures I outlined.
The second one can be reduced with the second procedure, but the first procedure won't work since the diagonal entries are not both positive. As a result you can only get what you call canonical form.

#### Sudharaka

##### Well-known member
MHB Math Helper
I interpret your problem statement as having a different basis for each of the quadratics.
The first one, which is positive definite, can be converted to normal form using the procedures I outlined.
The second one can be reduced with the second procedure, but the first procedure won't work since the diagonal entries are not both positive. As a result you can only get whay you call canonical form.
I think our definitions of the normal form is a little different. In our lecture notes the normal form is a representation of the quadratic form only in terms of squared terms (no cross terms) and all these squared terms should have coefficients +1, -1 or zero. I think your definition normal form doesn't contain -1 as a possible coefficient of the squared terms. #### Klaas van Aarsen

##### MHB Seeker
Staff member
I think our definitions of the normal form is a little different. In our lecture notes the normal form is a representation of the quadratic form only in terms of squared terms (no cross terms) and all these squared terms should have coefficients +1, -1 or zero. I think your definition normal form doesn't contain -1 as a possible coefficient of the squared terms. I have to admit that your references to normal form and canonical form for quadratics are unfamiliar to me.
What I do know, is that you can transform a quadratic q(x) into a normalized matrix form from which you can read off which conic section it is exactly, and you can also read off which transformations are necessary to transform it into a standard circle, parabola, hyperbola, or degenerate form. It seems to me that is exactly what your normal form is.

But they do have different axes.
Ah, then I guess you have to bring one in normal form using a coordinate transformation, and then express the other with the same transformations in such a way that it only contains squares (with transformation expressions between parentheses), which is what is apparently called the canonical form.
That should be doable.

#### Sudharaka

##### Well-known member
MHB Math Helper
I have to admit that your references to normal form and canonical form for quadratics are unfamiliar to me.
What I do know, is that you can transform a quadratic q(x) into a normalized matrix form from which you can read off which conic section it is exactly, and you can also read off which transformations are necessary to transform it into a standard circle, parabola, hyperbola, or degenerate form. It seems to me that is exactly what your normal form is.

But they do have different axes.
Ah, then I guess you have to bring one in normal form using a coordinate transformation, and then express the other with the same transformations in such a way that it only contains squares (with transformation expressions between parentheses), which is what is apparently called the canonical form.
That should be doable.
Exactly! But what makes you think it's doable and in fact how do we know whether this is doable? I got the thought that this might be impossible from the reasons I have given in my original post. #### Klaas van Aarsen

##### MHB Seeker
Staff member
Exactly! But what makes you think it's doable and in fact how do we know whether this is doable? I got the thought that this might be impossible from the reasons I have given in my original post. $q_1(x_1,\,x_2)=(x_1+x_2)^2+x_{2}^2=y^2+z^2$

$q_2(x_1,\,x_2)=(x_1+x_2)^2-5x_{2}^2=y^2-5z^2$
As it is, you have chosen your axes to be non-perpendicular to get to your normal form.
I guess that is okay.

I'd say that to get to canonical form for $q_2$ you would then need:
$q_2(x_1,\,x_2)=(x_1+x_2)^2-5x_{2}^2=y^2-(\sqrt 5 \cdot z)^2$

Doesn't that satisfy the criteria?

It shows that the second conic section is a hyperbola which is scaled in the z-direction with respect to the first conic section, which is an ellipse.

#### Sudharaka

##### Well-known member
MHB Math Helper
As it is, you have chosen your axes to be non-perpendicular to get to your normal form.
I guess that is okay.

I'd say that to get to canonical form for $q_2$ you would then need:
$q_2(x_1,\,x_2)=(x_1+x_2)^2-5x_{2}^2=y^2-(\sqrt 5 \cdot z)^2$

Doesn't that satisfy the criteria?

It shows that the second conic section is a hyperbola which is scaled in the z-direction with respect to the first conic section, which is an ellipse.
I think there's a little misunderstanding here. You have quoted the quadratic form that I have altered. The original ones given in the problem are,

$q_1(x_1,\,x_2)=x_{1}^{2}+2x_1 x_2+2x_{2}^{2}=(x_1+x_2)^2+x_{2}^2$

and

$q_2(x_1,\,x_2)=x_{1}^{2}-2x_1 x_2-4x_{2}^{2}=(x_1-x_2)^2-5x_{2}^2$

In my original post (in My Ideas section) I have altered the equation by putting a positive sign instead of a negative to show that there might be a typo in the question. #### Klaas van Aarsen

##### MHB Seeker
Staff member
I think there's a little misunderstanding here. You have quoted the quadratic form that I have altered. The original ones given in the problem are,

$q_1(x_1,\,x_2)=x_{1}^{2}+2x_1 x_2+2x_{2}^{2}=(x_1+x_2)^2+x_{2}^2$

and

$q_2(x_1,\,x_2)=x_{1}^{2}-2x_1 x_2-4x_{2}^{2}=(x_1-x_2)^2-5x_{2}^2$

In my original post (in My Ideas section) I have altered the equation by putting a positive sign instead of a negative to show that there might be a typo in the question. Sorry. My mistake.

Still, you have defined $y=x_1+x_2$ and $z=x_2$.
So you have $x_1=y-z$ and $x_2=z$.
Seems to me you can substitute that in the 2nd quadratic and normalize it, can't you?

I'm getting $q_2 =(y-2z)^2 - (\sqrt 5 \cdot z)^2$ .

#### Sudharaka

##### Well-known member
MHB Math Helper
Sorry. My mistake.

Still, you have defined $y=x_1+x_2$ and $z=x_2$.
So you have $x_1=y-z$ and $x_2=z$.
Seems to me you can substitute that in the 2nd quadratic and normalize it, can't you?

I'm getting $q_2 =(y-2z)^2 - (\sqrt 5 \cdot z)^2$ .
Exactly. But then the problem is that the second quadratic is normalized under a different transformation. It's normalized under the transformation,

$v=x_1-x_2\mbox{ and }w=\sqrt{5}x_2$

whereas the first quadratic is normalized under,

$y=x_1+x_2\mbox{ and }z=x_2$

As far as how I interpret the question it asks for a basis of $$V$$ where one of the quadratics is normalized and the other is canonical (the coefficients of the squared terms after the substitution can be any number).

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Exactly. But then the problem is that the second quadratic is normalized under a different transformation. It's normalized under the transformation,

$v=x_1-x_2\mbox{ and }w=\sqrt{5}x_2$

whereas the first quadratic is normalized under,

$y=x_1+x_2\mbox{ and }z=x_2$

As far as how I interpret the question it asks for a basis of $$V$$ where one of the quadratics is normalized and the other is canonical (the coefficients of the squared terms after the substitution can be any number).
Let me see if I understand you correctly.

If I do, you're looking for a solution of either:
\begin{array}{}
q_1(x_1, x_2) &=& \pm y^2 \pm z^2 &=& x_1^2 + 2x_1x_2 + 2x_2^2 \\
q_2(x_1, x_2) &=& a y^2 + bz^2 &=& x_1^2 - 2x_1x_2 - 4x_2^2
\end{array}
or
\begin{array}{}
q_1(x_1, x_2) &=& a y^2 + bz^2 &=& x_1^2 + 2x_1x_2 + 2x_2^2 \\
q_2(x_1, x_2) &=& \pm y^2 \pm z^2 &=& x_1^2 - 2x_1x_2 - 4x_2^2
\end{array}
where $y$ and $z$ are functions of $x_1$ and $x_2$.

Is that right?
If so, this is something we can try to solve...

Last edited:

#### Sudharaka

##### Well-known member
MHB Math Helper
Let me see if I understand you correctly.

If I do, you're looking for a solution of either:
\begin{array}{}
q_1(x_1, x_2) &=& \pm y^2 \pm z^2 &=& x_1^2 + 2x_1x_2 + 2x_2^2 \\
q_2(x_1, x_2) &=& a y^2 + b\color{red}{z}^2 &=& x_1^2 - 2x_1x_2 - 4x_2^2
\end{array}
or
\begin{array}{}
q_1(x_1, x_2) &=& a y^2 + b\color{red}{z}^2 &=& x_1^2 + 2x_1x_2 + 2x_2^2 \\
q_2(x_1, x_2) &=& \pm y^2 \pm z^2 &=& x_1^2 - 2x_1x_2 - 4x_2^2
\end{array}
where $y$ and $z$ are functions of $x_1$ and $x_2$.

Is that right?
If so, this is something we can try to solve...
Exactly. I guess you had some typos which I have corrected in red. #### Klaas van Aarsen

##### MHB Seeker
Staff member
Exactly. I guess you had some typos which I have corrected in red. Good!
Let me correct that in my post for reference, so I don't accidentally pick up the wrong formulas (again). There would be an extra condition: $y$ and $z$ have to be linear combinations of $x_1$ and $x_2$, since we're supposed to look for a basis.
If there is no such solution it can't be done.

Did you try to solve it?

#### Sudharaka

##### Well-known member
MHB Math Helper
Good!
Let me correct that in my post for reference, so I don't accidentally pick up the wrong formulas (again). There would be an extra condition: $y$ and $z$ have to be linear combinations of $x_1$ and $x_2$, since we're suppose to look for a basis.
If there is no such solution it can't be done.

Did you try to solve it?
Suppose you eliminate $$y$$ or $$z$$ using two equations and then subject the remaining variable. Then you'll end up with an ugly quadratic function of $$x_1$$ and $$x_2$$ inside the square root sign. And we'll have to find some value for $$a$$ and $$b$$ such that the quadratic has repeated roots so that the square root will cancel out. But how to find these values of $$a$$ and $$b$$ is the problem. I was confused by the complexity of this method that I abandoned it. #### Klaas van Aarsen

##### MHB Seeker
Staff member
Suppose you eliminate $$y$$ or $$z$$ using two equations and then subject the remaining variable. Then you'll end up with an ugly quadratic function of $$x_1$$ and $$x_2$$ inside the square root sign. And we'll have to find some value for $$a$$ and $$b$$ such that the quadratic has repeated roots so that the square root will cancel out. But how to find these values of $$a$$ and $$b$$ is the problem. I was confused by the complexity of this method that I abandoned it. I believe the "ugly" quadratic equation "proves" that there is no such linear solution, which should be provable by checking the linearity condition.
Not unexpected, since the 2 quadratics have different axes.

#### Sudharaka

##### Well-known member
MHB Math Helper
I believe the "ugly" quadratic equation "proves" that there is no such linear solution, which should be provable by checking the linearity condition.
Not unexpected, since the 2 quadratics have different axes.
Yes, checking the linearity condition is also difficult not knowing for which values of $$a$$ and $$b$$ it's linear. Anyway I think we don't need to waste more time on this question. It would be interesting what our prof would give as the solution. I'll post it here as soon as I have it. Thanks for all your help, I truly appreciate it. Although we didn't came up with an answer my understanding about many things improved due to this thread. #### Klaas van Aarsen

##### MHB Seeker
Staff member
Yes, checking the linearity condition is also difficult not knowing for which values of $$a$$ and $$b$$ it's linear. Anyway I think we don't need to waste more time on this question. It would be interesting what our prof would give as the solution. I'll post it here as soon as I have it. Thanks for all your help, I truly appreciate it. Although we didn't came up with an answer my understanding about many things improved due to this thread. It's not that hard.

Suppose we pick the case
\begin{array}{}
q_1(x_1, x_2) &=& y^2 + z^2 &=& x_1^2 + 2x_1x_2 + 2x_2^2 \\
q_2(x_1, x_2) &=& a y^2 + bz^2 &=& x_1^2 - 2x_1x_2 - 4x_2^2
\end{array}

Then we get that
$$z=\sqrt{\frac{(a-1)x_1^2+2(a+1)x_1x_2+2(a+2)x_2^2}{a-b}}$$
where a has to be different from b, or we won't have a viable solution.
This has to hold true for any $x_1$ and any $x_2$ with the extra condition that if we multiply say $x_1$ by $\lambda$, that $z$ is also multiplied by $\lambda$.
Since there is at least one non-zero term within the square root that is not a square, this is not possible.
The same argument holds for all other choices.
QED

#### Sudharaka

##### Well-known member
MHB Math Helper
It's not that hard.

Suppose we pick the case
\begin{array}{}
q_1(x_1, x_2) &=& y^2 + z^2 &=& x_1^2 + 2x_1x_2 + 2x_2^2 \\
q_2(x_1, x_2) &=& a y^2 + bz^2 &=& x_1^2 - 2x_1x_2 - 4x_2^2
\end{array}

Then we get that
$$z=\sqrt{\frac{(a-1)x_1^2+2(a+1)x_1x_2+2(a+2)x_2^2}{a-b}}$$
where a has to be different from b, or we won't have a viable solution.
This has to hold true for any $x_1$ and any $x_2$ with the extra condition that if we multiply say $x_1$ by $\lambda$, that $z$ is also multiplied by $\lambda$.
Since there is at least one non-zero term within the square root that is not a square, this is not possible.
The same argument holds for all other choices.
QED
Yes this makes perfect sense of why this is not possible. I suspect our prof meant something else in this question, but we'll have to wait and see what it is. Anyway thanks again for all your help. I am having a hard time with Linear Algebra these days. #### Klaas van Aarsen

##### MHB Seeker
Staff member
I am having a hard time with Linear Algebra these days. Seems to me that you're picking it up quickly and completely.
You shouldn't have any problems when you get to any examinations!

#### Sudharaka

##### Well-known member
MHB Math Helper
Seems to me that you're picking it up quickly and completely.
You shouldn't have any problems when you get to any examinations!
I don't know..... All the assignments and the exams for the course are really tuff and I always feel that I don't have the necessary prerequisites needed for the course. That's why I ask many questions about Linear Algebra these days.