# $T:\Bbb R^2 \to \Bbb R^2$, linear, diagonal with respect to any basis.

Is there a linear transformation from $\Bbb R^2$ to $\Bbb R^2$ which is represented by a diagonal matrix when written with respect to any fixed basis?

If such linear transformation $T$ exists, then its eigenvector should be the identity matrix for any fixed basis $\beta$ of $\Bbb R^2$.
Then, I don't see, if this is possible or not.

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Since $T$ is diagonal in at least one basis, $T$ has two eigenvalues $a$ and $b$ and two linearly independent eigenvectors $x$ and $y$ such that $Tx=ax$ and $Ty=by$.

If $a\ne b$, $x+y$ is not an eigenvector hence the matrix of $T$ in the basis $(x+y,y)$ is not diagonal. This is absurd. Hence, $a=b$, that is, $T=aI$.

In the other direction, if $T=aI$ for some $a$, then its matrix in any basis is diagonal.

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Way to go, Did. +1 – 1015 Feb 7 '13 at 20:20

Yes: More generally: let $R$ be a unital ring, then the for a matrix $Z\in M_{n\times n}(R)$ the following are equivalent:

1. is invariant with respect to any chosen basis of $R^n$
2. $Z$ is representable as a diagonal matrix in some basis of $R^n$.
3. $Z$ is a scalar matrix, ie: $\left( (\exists r \in R) Z=rI_n \right)$
4. $Z$ is in the center of $GL_n(R)$.

Proof (assuming 4 $\Leftrightarrow 2$ which can be found in a text on algebraic groups, or affine group schemes):

Any element $Z$ of the center of $M_{n\times n}(R)$ then satisfies:

$$(\forall P \in GL_n(R)) \mbox{ } Z= IZ=PP^{-1}Z = PZ P^{-1}.$$

However, since the center of $M_{n\times n}(R)$ is the diagonal subgroup of $GL_n(R)$ and since any change of basis of $R^n$ is the same as conjugating by some $P\in GL_n(R)$.

Therefore a matrix is the same under any basis of $R^n$ if and only if it is invariant with respect to any conjugation by some $P\in GL_n(R)$ if and only if it is in the center of $GL_n(R)$ if and only if it is diagonal is some basis if and only if its is a scalar matrix.

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If the transformation $T$ is represented by the matrix $A$ in basis $\mathcal{A}$, then it is represented by the matrix $PAP^{-1}$ in basis $\mathcal{B}$, where $P$ is the invertible change-of-basis matrix.

Suppose that $T$ is represented by a diagonal matrix in any basis. Let $P$ be an arbitrary invertible matrix and $A$ any diagonal matrix:

$$P = \left[\begin{array}{cc} p_{1,1} & p_{1,2} \\ p_{2,1} & p_{2,2} \end{array}\right] \text{ and } A = \left[\begin{array}{cc} d_1 & 0 \\ 0 & d_2 \end{array}\right].$$

Now, calculate

$PAP^{-1} = \dfrac{1}{\det P} \left[\begin{array}{cc} b_{1,1} & b_{1,2} \\ b_{2,1} & b_{2,2} \end{array}\right]$, where the entries $b_{i,j}$ are polynomials in the $p_{i,j}$ and $d_i$ variables.

For this new conjugated matrix to be diagonal, we have the following two equations. (Check!)

\begin{align*} 0 = b_{1,2} &= (d_2 - d_1)p_{1,1}p_{1,2} \\ 0 = b_{2,1} &= (d_1 - d_2)p_{2,1}p_{2,2} \end{align*}

Since $P$ is arbitrary, the only way for these equations to always be satisfied is for $d_1 = d_2$. In other words, the original matrix $A$ was a scalar multiple of the identity.

$$A = d \cdot \operatorname{Id}_2 = \left[\begin{array}{cc} d & 0 \\ 0 & d \end{array}\right].$$

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If $T$ is diagonal with respect to any basis, then every nonzero vector is an eigenvector, since any nonzero vector can be extended to a basis. Then it must be that the whole vector space is one eigenspace corresponding to the same eigenvalue, since if two vectors were eigenvectors with different eigenvalues, then the sum would not be an eigenvector. Since any matrix acts on an eigenspace $E$ as $\lambda I$, then it follows $T = \lambda I$, where $\lambda$ is some scalar.

So the identity matrix works, but so does any multiple of the identity.

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How do you prove your first claim? – 1015 Feb 7 '13 at 20:20
Every vector can be extended to a basis, and by assumption every basis is a basis of eigenvectors. – Christopher A. Wong Feb 7 '13 at 20:25
Yes. Maybe you could add this sentence to your answer? (every nonzero vector, btw) Just a suggestion, of course. Note that "then it must be" is not transparent for everyone either. – 1015 Feb 7 '13 at 20:30
This is certainly more elegant than my answer, but I figured a pedestrian calculation would be nice to see, too. – Sammy Black Feb 7 '13 at 20:39
Thanks for the edit. – 1015 Feb 7 '13 at 20:54

Yes: it's the identity transformation $Tx = x$.

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What about T=42.I? – Did Feb 7 '13 at 20:13