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Let $M_{2} (\Bbb R)$ denote the set of $2 \times 2$ matrices. Let $ A \in M_{2} (\Bbb R)$ be of trace $2$ and determinant $-3$. Identifying $M_{2} (\Bbb R)$ as $\Bbb R^4$, consider a linear transformation $ T : M_{2} (\Bbb R)\rightarrow$$M_{2} (\Bbb R)$ defined by $T(B)= AB$. Then which of the following hold:

  1. $T$ is diagonalizable.
  2. $T$ is invertible.
  3. $2$ is an eigenvalue of $T$.
  4. $T(B)=B$ for some $B\neq 0$.

Please suggest which of the options are correct. It seems that $3$ and $-1$ are the eigen values for $A$ so (1) holds and (2) does not.

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What do you know about matrices with non-zero determinant? – Brian M. Scott May 13 '12 at 19:13
Here's a huge hint: Construct dyadic matrices from the eigenvectors of $A$. – copper.hat May 13 '12 at 19:35
HINT: Let $v_1,v_2$ be orthonormal eigenvectors of $A$ corresponding to $A$'s distinct eigenvalues. Now compute $T(v_i v_j^T)$, and think about the answer. – copper.hat May 14 '12 at 0:02
@copper.hat: Thanks for the hint. But can you please illustrate it bit more. – preeti May 14 '12 at 18:10
Did you compute $T(v_i v_j^T)$? – copper.hat May 14 '12 at 18:13
up vote 3 down vote accepted

Fixed to be about $T$, not about $A$.

The trace of a square matrix equals the sum of the eigenvalues (in the algebraic closure of the ground field, if necessary; i.e., the sum of the roots of the characteristic polynomial); the determinant equals the product of the eigenvalues. This often gives a nice way of finding at least some of the eigenvalues, and in the case of $2\times 2$ matrix, gives all the information required to find all the eigenvalues (since knowing $a+b$ and $ab$ will determine $a$ and $b$).

Here you have a $2\times 2$ matrix, so it has two eigenvalues; their sum is $2$ and their product is $-3$. Thus, they are $3$ and $-1$.

In particular, $A$ is invertible, since no eigenvalue is equal to $0$.

Now let's consider $T$. Note that $T$ is one-to-one, because if $T(B)=0$, then $AB=0$. But since $A$ is invertible, this means that $B=A^{-1}AB=A^{-1}0= 0$.

Thus, $T$ is one-to-one on a finite dimensional vector space, so $T$ is invertible. This proves that (2) is true.

Now, notice that if $AB=\lambda B$, then $A\mathbf{b}_i=\lambda\mathbf{b}_i$, where $\mathbf{b}_i$ is the $i$th column of $A$, since $$AB = A(\mathbf{b}_1\;\mathbf{b}_2) = (A\mathbf{b}_1\;A\mathbf{b}_2).$$ In particular, if $B$ is an eigenvector of $T$, then $B\neq 0$, and therefore either $\mathbf{b}_1$ or $\mathbf{b}_2$ are nonzero, so either $A\mathbf{b}_1=\lambda\mathbf{b}_1$ or $A\mathbf{b}_2=\lambda\mathbf{b}_2$ shows that $\lambda$ is an eigenvalue of $A$. Conversely, if both $\mathbf{b}_1$ and $\mathbf{b}_2$ are in the eigenspace of $\lambda$ for $A$, and they are not both zero, then $B=(\mathbf{b}_1\;\mathbf{b}_2)$ is an eigenvector of $T$ associated to $\lambda$. That means that the only possible eigenvalues of $T$ are the eigenvalues of $A$. This proves that both (2) and (4) are false, since neither $2$ nor $1$ are eigenvalues of $A$, so they are not eigenvalues of $T$.

The only thing left is whether $T$ is diagonalizable.

As noted above, a matrix $B$ is an eigenvector of $T$ associated to $\lambda$ if and only if both columns of $B$ lie in the eigenspace of $A$ associated to $\lambda$. Since the eigenspaces of $A$ are one-dimensional, we can select $\mathbf{v}_1$ and $\mathbf{v}_2$, eigenvectors of $A$ associated to $3$ and $-1$, respectively. Then $B$ is an eigenvector of $T$ associated to $3$ if and only if $B=(\alpha\mathbf{v}_1\;\beta\mathbf{v}_2)$ and $\alpha$ and $\beta$ are not both zero. Thus, we have two degrees of freedom, so the eigenspace of $T$ associated to $3$ has dimension $2$. Similarly, the eigenvectors of $T$ associated to $-1$ are of the form $B=(\rho\mathbf{v}_2\;\sigma\mathbf{v}_2)$ with $\rho$ and $\sigma$ arbitrary but not both zero; again, the dimension is $2$.

Since the sum of the geometric dimensions of the eigenspaces of $T$ is $4$, which is the dimension of the vector space $M_2(R)$, this proves that $T$ is diagonalizable. So (1) is true.

In summary, (1) and (3) are true, (2) and (4) are false.

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You seem to be answering the question about $A$ rather than $T$. $T$ does not have distinct eigenvalues. – Ted May 13 '12 at 23:59
There is a difference between $A$ and $T$. $T$ has 4, not 2, eigenvalues. – copper.hat May 14 '12 at 0:00
Oops; quite so. Sorry about that. – Arturo Magidin May 14 '12 at 0:00

Hint 1:

If $$ A=\begin{bmatrix}a&b\\c&d\end{bmatrix} $$ then using the equation $T(B)=AB$, we get the following matrix for $T$ $$ T=\begin{bmatrix}a&0&b&0\\0&a&0&b\\c&0&d&0\\0&c&0&d\end{bmatrix} $$ Hint 2:

$\det(T)=(ad-bc)^2=\det(A)^2$. Thus, replacing $a$ and $d$ by $a-x$ and $d-x$ we get that the characteristic polynomial of $T$ is the square of the characteristic polynomial for $A$. That is $(x^2-2x-3)^2=(x-3)^2(x+1)^2$.

Hint 3:

Since $A$ is diagonalizable (it has distinct eigenvalues), it has two eigenvectors which span $\mathbb{R}^2$: $(x_1,y_1)$ and $(x_2,y_2)$. $(x_1,0,y_1,0), (0,x_1,0,y_1), (x_2,0,y_2,0), (0,x_2,0,y_2)$ are eigenvectors of $T$.

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Note that for $2\times 2$ matrices, we have the simple characteristic polynomial $$p(x) = x^2 - tr(A)x + det(A) = x^2 - 2x - 3 = (x - 3)(x + 1)$$ so the eigenvalues are indeed $3$ and $-1$.

  1. What can you say about matrices which have distinct eigenvalues? Think in terms of the geometric multiplicity and algebraic multiplicity.

  2. 2 is not an eigenvalue, as we have just shown above.

  3. The determinant is -3 and hence non-zero. What can you say about matrices with non-zero determinant?

  4. Think about what the equation $T(B) = B$ represents. We can split the columns of $$B = \begin{bmatrix} \vec{v_1} & \vec{v_2} \end{bmatrix}$$ to get the equations $$T\vec{v_1} = \vec{v_1}$$ and also $$T\vec{v_2} = \vec{v_2}$$ What can you tell me then about the relationship between the vectors and $T$? Is it possible for them to exist?

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You seem to be answering the questions about $A$, but the questions are about $T$. – Ted May 13 '12 at 19:26
You may be right, but you haven't shown that 2 is not an eigenvalue of $T$. – copper.hat May 13 '12 at 19:32
@copper.hat The invariants of $T$ can be computed from any matrix representing $T$. If he has shown that the eigenvalues of $T$ are -3 and 1, how could it also have 2 as an eigenvalue? – rschwieb May 13 '12 at 21:23
@rschwieb: I did not say that 2 could be an eigenvalue. $T$ has 4 eigenvalues, the above has a degree 2 polynomial. In the above, $T$ is operating on $\mathbb{R}^2$. Some connection is missing. The characteristic polynomial of $T$ is $p^2$, not $p$. – copper.hat May 13 '12 at 22:22
@copper.hat Ah I see what the two of you are talking about now. I hadn't read the unusual setup for this problem carefully enough. – rschwieb May 13 '12 at 22:30

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