# Questions about a linear operator induced by a diagonal matrix

Let $A \in \mathbb{C}^{4 \times 4}$ be a diagonal matrix with main diagonal entries $1,2,3,4$. Define $T_A : \mathbb{C}^{4 \times 4} \rightarrow \mathbb{C}^{4 \times 4}$ by $T_A(B) = AB-BA$.

I am studying for a test and trying to answer some questions about $T_A$ but the answers I have come up with are very computational and I was wondering if there is a faster way to attack this problem then the method I will describe.

Part 1) What are $\dim(\ker(T_A))$ and $\dim(Image(T_A))$

I was thinking there might be some slick way to use the rank nullity theorem without actually having to do computation but so far my only answer involves calculating the matrix expressions:

$$AB = \left( \begin{array}{ccc} b_{11} & \ldots & b_{14} \\ 2b_{21} & \ldots & 2b_{24} \\ 3b_{31} & \ldots & 3b_{34} \\ 4b_{41} & \ldots & 4b_{44} \end{array} \right), BA = \left( \begin{array}{ccc} b_{11} & 2b_{12} & 3b_{13} & 4b_{14} \\ b_{11} & 2b_{12} & 3b_{13} & 4b_{14} \\ b_{11} & 2b_{12} & 3b_{13} & 4b_{14} \\ b_{11} & 2b_{12} & 3b_{13} & 4b_{14} \end{array} \right)$$

To calculate the dimension of the kernel we just set $AB = BA$ and observe $2b_{21} = b{21}$ so $b_{21} =0$ and comparing the rest of the entries the same way we get $B \in \ker(T_A)$ when $B = \left( \begin{array}{ccc} b_{11} & 0 & 0 & 0 \\ 0 & 2b_{22} & 0 & 0 \\ 0 & 0 & 3b_{33} & 0 \\ 0 & 0 & 0 & 4b_{44} \end{array} \right)$

Therefore $\dim(\ker(T_A)) = 4$

Is there a faster way to do this?

Using $\dim(ker(T_A)) + \dim(Image(T_A)) = \dim(V)$

we can now see $4+ Image(T_A) = 16$ therefore $Image(T_A) = 12$

part 2) What are the eigenvalues of $T_A$

Using the equation $AB-BA = \lambda B$ and the calculations from part 1 implies

$$\left( \begin{array}{ccc} b_{11} & \ldots & b_{14} \\ 2b_{21} & \ldots & 2b_{24} \\ 3b_{31} & \ldots & 3b_{34} \\ 4b_{41} & \ldots & 4b_{44} \end{array} \right) - \left( \begin{array}{ccc} b_{11} & 2b_{12} & 3b_{13} & 4b_{14} \\ b_{21} & 2b_{22} & 3b_{23} & 4b_{24} \\ b_{31} & 2b_{32} & 3b_{33} & 4b_{34} \\ b_{41} & 2b_{42} & 3b_{43} & 4b_{44} \end{array} \right) = \left( \begin{array}{ccc} 0 & -b_{12} & -2b_{13} & -3b_{14} \\ b_{21} & 0 & -b_{23} & -2b_{24} \\ 2b_{31} & b_{32} & 0 & -b_{34} \\ 3b_{41} & 2b_{42} & b_{43} & 0 \end{array} \right)$$

equating this with $\lambda B$ implies $B = \left( \begin{array}{ccc} 0 & 0 & 0 & 0 \\ b_{21} & 0 & 0 & 0 \\ 0 & b_{32} & 0 & 0 \\ 0 & 0 & b_{43} & 0 \end{array} \right)$

Which gives us the eigenvectors of $T_A$ with four repeated eigenvalues $\lambda =1$

We see the characteristic polynomial of $T_A = (x-1)^4$

Question 1) Is there a simple way to do any of part1) or part2) without resorting to all the matrix calculations. In particular I also want to compute the minimal polynomial of $T_A$ without having to compute powers of $(AB-BA-I)$.

Question2) Using the structure of the eigenvectors obtained in part 2) can we list a basis in which $T_A$ is diagonalizable?

• Perhaps it's just little tricks. It seems a computational question. You can look at basis elements for your space - there are 16 of them, each with fifteen 0s and one 1. The diagonal ones commute with A (in fact, diagonal matrices always commute). To see that the others don't: left multiplication by an invertible matrix is a row operation, and right multiplication is a column operation. This one's easy: $B\mapsto AB$ multiplies the first row by 1, the second by 2, etc. So the others won't commute (try it). For part 2, your answer looks like just about the only sensible thing to me. Sorry! Aug 1, 2011 at 18:11
• Oh, except you've only found the eigenvectors corresponding to the value 1. There are eigenvectors of eigenvalue 2, too... Aug 1, 2011 at 18:14
• Billy's basis is the one to use. If $E_{ij}$ is the matrix with 1 in the $(i,j)$ position, $T_A(E_{ij}) = (i-j) E_{ij}$. These are all the eigenvectors you need. Aug 1, 2011 at 18:37
• Does this show that $T_A$ is diagonalizable? Aug 1, 2011 at 20:03

Your answer to part 1 is fine, but in part 2, since $T_A$ is a linear mapping defined on $\mathbb{C}^{4\times 4}$, which is a vector space of dimension 16, the characteristic polynomial should be of degree 16, not a degree-4 polynomial.

There are other ways to solve your problem, but let us follow your approach first. To find the eigenvalues of $T_A$, we inspect the equation $T_A(B) \equiv AB-BA = \lambda B$ and we obtain $$(*): \begin{pmatrix} 0 & -b_{12} & -2b_{13} & -3b_{14}\\ b_{21} & 0 & -b_{23} & -2b_{24}\\ 2b_{31} & b_{32} & 0 & -3b_{14}\\ 3b_{41} & 2b_{42} & b_{43} & 0 \end{pmatrix} = \begin{pmatrix} \lambda b_{11} & \lambda b_{12} & \lambda b_{13} & \lambda b_{14}\\ \lambda b_{21} & \lambda b_{22} & \lambda b_{23} & \lambda b_{24}\\ \lambda b_{31} & \lambda b_{32} & \lambda b_{33} & \lambda b_{14}\\ \lambda b_{41} & \lambda b_{42} & \lambda b_{43} & \lambda b_{44} \end{pmatrix}.$$ The above equation actually have seven different roots, not one. These seven roots are $\lambda = -3, -2, -1, 0, 1, 2, 3$. For instance, when $\lambda = 3$, the corresponding "eigenvectors" are the matrices of the following form: $$B = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ b_{41} & 0 & 0 & 0 \end{pmatrix}.$$ (You may verify that such a matrix $B$ will satisfy $(*)$ for $\lambda=3$.) So the eigenspace for $\lambda=3$ has dimension 1, i.e. $\lambda=3$ contributes the factor $(x-3)$ to the characteristic polynomial.

Another example. For $\lambda=0$, all matrices $B$ that satisfy $(*)$ are of the form $$\begin{pmatrix} b_{11} & 0 & 0 & 0\\ 0 & b_{22} & 0 & 0\\ 0 & 0 & b_{33} & 0\\ 0 & 0 & 0 & b_{44}\\ \end{pmatrix}.$$ Hence the eigenspace has dimension 4, i.e. the eigenvalue $\lambda=0$ contributes the factor $x^4$ to the characteristic polynomial.

The eigenvectors for the other eigenvalues can be found in a similar manner. In summary, the characteristic polynomial of $T_A$ should be $p(x)=(x+3)(x+2)^2(x+1)^3x^4(x-1)^3(x-2)^2(x-3)$, or equivalently, $p(x)=x^4(x^2-1)^3(x^2-4)^2(x^2-9)$. And the minimal polynomial is $m(x)=x(x^2-1)(x^2-4)(x^2-9)$.

One thing that may be confusing is that the "eigenvectors" in this problem are actually square matrices. This is unlike the usual case, where the "vectors" we talk about are genuine column vectors. Yet your problem can be recast to the familiar form, but you need the knowledge of Kronecker product (a.k.a. tensor product). You also need to turn the $4\times4$ matrix $B$ into a 16-vector ${\rm vec}(B)$. This can be done by stacking up the columns of $B$, i.e. we represent $B$ by the following column vector: $${\rm vec}(B)=(b_{11}, b_{21}, b_{31}, b_{41}, b_{12}, b_{22}, b_{32}, b_{42}, b_{13}, b_{23}, b_{33}, b_{43}, b_{14}, b_{24}, b_{34}, b_{44})^T.$$ (Note the transpose at the end of the above, so ${\rm vec}(B)$ is a column vector, not a row vector.) In general, for a linear mapping of the form $T(B)=ABC$, we have ${\rm vec}(T(B)) = (C^T\otimes A){\rm vec}(B)$. In other words, if we represent both $B$ and $T(B)$ in vector forms (instead of in matrix forms), then the matrix representation of $T$ is given by $C^T\otimes A$.

In our case, the equation $AB-BA=\lambda B$ can be rewritten as $ABI-IBA=\lambda B$, or $(I\otimes A - A^T\otimes I){\rm vec}(B) = \lambda {\rm vec}(B)$, where $(I\otimes A - A^T\otimes I)$ is a $16\times16$ matrix and ${\rm vec}(B)$ is a column vector with 16 entries. Since our $A$ here is a diagonal matrix, writing down $(I\otimes A - A^T\otimes I)$ is especially easy -- $I\otimes A$ is the diagonal matrix with diagonal $$d_1 = (1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4),$$ whereas $A^T\otimes I$ is the diagonal matrix with diagonal $$d_2 = (1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4).$$ Therefore, the matrix representation of $T_A$, i.e. $(I\otimes A - A^T\otimes I)$, is the $16\times16$ diagonal matrix with diagonal $$d_1 - d_2 = (0, 1, 2, 3, -1, 0, 1, 2, -2, -1, 0, 1, -3, -2, -1, 0).$$ Solving the eigenvalue problem $(I\otimes A - A^T\otimes I)v = \lambda v$ in this case is just a piece of cake, despite it is a 16-dimensional problem.

I hope this is concrete enough. For a more "abstract" or "algebraic" method, see Robert Israel's comment in the above.

A final note: in the above characteristic polynomial and minimal polynomial, you may substitute for $x$ the matrix representation of $T_A$, i.e. $(I\otimes A - A^T\otimes I)$, to obtain a zero matrix of order 16.

• Did you find the minimal polynomial without computing $m(T_A)$? Aug 1, 2011 at 20:02
• Yes. For an eigenvalue problem $Mv = \lambda v$, if $M$ is diagonalizable and $\lambda_1, ..., \lambda_k$ are the distinct eigenvalues of $M$, the minimal polynomial is simply $m(x)=(x-\lambda_1)...(x-\lambda_k)$. That is, each different eigenvalue only contributes a linear factor to the minimal polynomial. This is a general truth. When $M$ is not diagonalizable, things get nastier, but in our case, $M=(I\otimes A - A^T\otimes I)$ is already a diagonal matrix. Aug 1, 2011 at 23:06