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[SOLVED] Canonical Basis and Standard Basis

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hi everyone, :)

I have a little trouble understanding what Canonical basis means in the following question. I thought that Canonical basis is just another word for the Standard basis. Hope you people could clarify the difference between these two in the given context. :)

Question:

Find the canonical basis for the orthogonal thransformation \(f:\Re^3\rightarrow \Re^3\) such that \(A_{f,\,B}=\frac{1}{3}\begin{pmatrix}2&-1&2\\2&2&-1\\-1&2&2\end{pmatrix}\), \(B\) being a standard basis of \(\Re^3\).
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Hi everyone, :)

I have a little trouble understanding what Canonical basis means in the following question. I thought that Canonical basis is just another word for the Standard basis. Hope you people could clarify the difference between these two in the given context. :)

Question:

Find the canonical basis for the orthogonal thransformation \(f:\Re^3\rightarrow \Re^3\) such that \(A_{f,\,B}=\frac{1}{3}\begin{pmatrix}2&-1&2\\2&2&-1\\-1&2&2\end{pmatrix}\), \(B\) being a standard basis of \(\Re^3\).
I'm not sure either, but the only thing I imagine it could mean is the image of the standard basis under the given transformation. I.e. if you have a linear transformation $A : \mathbb{R}^3 \to \mathbb{R}^3$, then the canonical basis for this transformation is:

$$\left \{ A \left [ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right ], A \left [ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right ], A \left [ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right ] \right \}$$

(so it is the standard basis in the coordinate system described by this linear transformation)
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
I'm not sure either, but the only thing I imagine it could mean is the image of the standard basis under the given transformation. I.e. if you have a linear transformation $A : \mathbb{R}^3 \to \mathbb{R}^3$, then the canonical basis for this transformation is:

$$\left \{ A \left [ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right ], A \left [ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right ], A \left [ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right ] \right \}$$

(so it is the standard basis in the coordinate system described by this linear transformation)
Thanks very much for the answer. I guess so too. But the thing is, for a grad level assignment this is too easy which keeps me thinking whether this is what our prof meant. :)
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Thanks very much for the answer. I guess so too. But the thing is, for a grad level assignment this is too easy which keeps me thinking whether this is what our prof meant. :)
Yes, that was my thought as well. It likely means something a bit less trivial, but I haven't seen this used anywhere and I can't find anything like it in my linear algebra book... so this was my best guess. Looking forward to other replies (Star)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
The word canonical means obvious. Exactly that.
A canonical basis might be the standard basis, but that is unlikely, since they would have just called it the standard basis then.

If the matrix were diagonalizable with real numbers, I would expect it to be the normalized basis of eigenvectors.
However, your matrix is only diagonalizable with complex numbers.
Can it be that you're supposed to diagonalize it with complex numbers?
It would fit with the other problems you have brought here.
 

smile

New member
Oct 15, 2013
19
Hi everyone, :)

I have a little trouble understanding what Canonical basis means in the following question. I thought that Canonical basis is just another word for the Standard basis. Hope you people could clarify the difference between these two in the given context. :)

Question:

Find the canonical basis for the orthogonal thransformation \(f:\Re^3\rightarrow \Re^3\) such that \(A_{f,\,B}=\frac{1}{3}\begin{pmatrix}2&-1&2\\2&2&-1\\-1&2&2\end{pmatrix}\), \(B\) being a standard basis of \(\Re^3\).
I think we need to find the eigenvectors for this matrix, then we can use them to diagonalize this matrix. But the problem is that I am not sure wether we need to normalize those eigenvectors or not. If we do, it is not easy to normalize them.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
The word canonical means obvious. Exactly that.
A canonical basis might be the standard basis, but that is unlikely, since they would have just called it the standard basis then.

If the matrix were diagonalizable with real numbers, I would expect it to be the normalized basis of eigenvectors.
However, your matrix is only diagonalizable with complex numbers.
Can it be that you're supposed to diagonalize it with complex numbers?
It would fit with the other problems you have brought here.
Yeah, I think for the moment this is the best assumption that we could make. However the problem with this is that in the problem we are dealing with the real number field and diagonalizing the matrix in complex number field would make our canonical basis not so obvious. Isn't? :p

I think we need to find the eigenvectors for this matrix, then we can use them to diagonalize this matrix. But the problem is that I am not sure wether we need to normalize those eigenvectors or not. If we do, it is not easy to normalize them.
I think we need to normalize the eigenvectors. The reason is otherwise it won't be unique. Am I correct ILSe? :)
 

smile

New member
Oct 15, 2013
19
Yeah, I think for the moment this is the best assumption that we could make. However the problem with this is that in the problem we are dealing with the real number field and diagonalizing the matrix in complex number field would make our canonical basis not so obvious. Isn't? :p



I think we need to normalize the eigenvectors. The reason is otherwise it won't be unique. Am I correct ILSe? :)
Yes, you are right, actually the normalization is not hard.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I think we need to normalize the eigenvectors. The reason is otherwise it won't be unique. Am I correct ILSe? :)
Since they ask for a canonical basis, I think you can get away with it without normalizing it.
You'd have a canonical basis of eigenvectors.
If they wanted it to be normalized, they should have asked for a normalized basis.

Either way, the problem statement is ambiguous, which is bad in math.
They should know better in my opinion.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Since they ask for a canonical basis, I think you can get away with it without normalizing it.
You'd have a canonical basis of eigenvectors.
If they wanted it to be normalized, they should have asked for a normalized basis.

Either way, the problem statement is ambiguous, which is bad in math.
They should know better in my opinion.
Yes, that was my thought as well. It likely means something a bit less trivial, but I haven't seen this used anywhere and I can't find anything like it in my linear algebra book... so this was my best guess. Looking forward to other replies (Star)
Thank you ILSe and Bacterius for all the valuable insight that you have provided on this question. It immensely helped me to clarify my doubts. :)