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matrix equation

jacks

Well-known member
Apr 5, 2012
226
Find all square matricx $A$ that satisfy the equation $A^2-4A+3I=0$

where $I$ is a Identity matrix
 

chisigma

Well-known member
Feb 13, 2012
1,704
Find all square matricx $A$ that satisfy the equation $A^2-4A+3I=0$

where $I$ is a Identity matrix
Is...

$\displaystyle A^{2}- 4 A +3 I= A\ (A -4\ I + 3\ A^{-1})= A\ (A -3\ I - I + 3\ A^{-1})= A\ \{(A-3\ I)- A^{-1}\ (A-3\ I)\}= A\ (A-3\ I)\ (I-A^{-1})$

... so that the A satisfying the given equation also satisfy one of the equations...

$\displaystyle A-3\ I=0$

$\displaystyle A-I=0$

Kind regards

$\chi$ $\sigma$
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Is...

$\displaystyle A^{2}- 4 A +3 I= A\ (A -4\ I + 3\ A^{-1})= A\ (A -3\ I - I + 3\ A^{-1})= A\ \{(A-3\ I)- A^{-1}\ (A-3\ I)\}= A\ (A-3\ I)\ (I-A^{-1})$

... so that the A satisfying the given equation also satisfy one of the equations...

$\displaystyle A-3\ I=0$

$\displaystyle A-I=0$

Kind regards

$\chi$ $\sigma$
The quadratic in A can be factorised straight off without having to worry about the existance of inverses. But I'm not sure that we can conclude that if a product is zero then one of the terms in the product is zero in this case.

CB
 

chisigma

Well-known member
Feb 13, 2012
1,704
The quadratic in A can be factorised straight off without having to worry about the existance of inverses. But I'm not sure that we can conclude that if a product is zero then one of the terms in the product is zero in this case.

CB
A trivial test confirms that $A=I$ and $A=3\ I$ are solutions of the equation...

$\displaystyle A^{2}-4\ A+3\ I=0$ (1)

The question is: there are some other solutions to (1)?... it is clear that the question is strictly connected to another question: can exist two matrices X and Y, with $X\ne 0$ and $Y\ne 0$, for whose is $X\ Y=0$?... a very interesting question!...

Kind regards

$\chi$ $\sigma$
 

CaptainBlack

Well-known member
Jan 26, 2012
890
A trivial test confirms that $A=I$ and $A=3\ I$ are solutions of the equation...

$\displaystyle A^{2}-4\ A+3\ I=0$ (1)

The question is: there are some other solutions to (1)?... it is clear that the question is strictly connected to another question: can exist two matrices X and Y, with $X\ne 0$ and $Y\ne 0$, for whose is $X\ Y=0$?... a very interesting question!...

Kind regards

$\chi$ $\sigma$
The answer to the last question is yes:
\[ A=\left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right] \]
Then \( A^2=0 \)

CB
 

jacks

Well-known member
Apr 5, 2012
226
Thanks to all experts

but my doubt is , is there is any other matrix exists for which $A=\begin{bmatrix}a &b \\
c & d
\end{bmatrix}$ which satisfy $A^2-4A+3I=0$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Thanks to all experts

but my doubt is , is there is any other matrix exists for which $A=\begin{bmatrix}a &b \\
c & d
\end{bmatrix}$ which satisfy $A^2-4A+3I=0$
Yes.
$$
A = \begin{bmatrix}3&0\\0&1\end{bmatrix}
$$

I just used the Cayley-Hamilton Theorem in reverse.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Yes.
$$
A = \begin{bmatrix}3&0\\0&1\end{bmatrix}
$$

I just used the Cayley-Hamilton Theorem in reverse.
Which is that a matrix \(A\) satisfies its' own characteristics equation:

\[\det (A-\lambda I)=0\]

Which off course allows you to generate many 2x2 solutions of the original equation.

But is that (with the two we already have, which are not restricted to being 2x2) all of them?

CB
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Which is that a matrix \(A\) satisfies its' own characteristics equation:

\[\det (A-\lambda I)=0\]

Which off course allows you to generate many 2x2 solutions of the original equation.

But is that (with the two we already have, which are not restricted to being 2x2) all of them?

CB
OK, I know it is bad form to reply to ones own posts, but:

Let us restrict our attention to 2x2 matrices. We know that \(A\) satisfies its' charateristic equation, which we may write as: \(\lambda^2+b\lambda+c=0\) and so \(A^2+bA+cI=0\), now suppose it also satisfies another second degree equation: \(A^2+\beta A+\gamma I =0\) (which forces \(\beta \ne b\) if these are to be different and consistent).

Then we have: \((b-\beta)A+(c-\gamma)I=0\), and so: \(A=\frac{\gamma-c}{b-\beta}I\)

So the 2x2 solutions to \(A^2-4A+3I=0 \) are either matrices with charateristic equation \(\lambda^2-4\lambda+3=0\) or are multiples of the identity (and so of neccessity the solutions we found by factorising the equation)

CB
 
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