# Eigenvalues of a matrix $A$ and Linear Tranformation

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$$M_{2} (\Bbb R) defined by T(B)= AB. Then which of the following hold: 1. T is diagonalizable. 2. T is invertible. 3. 2 is an eigenvalue of T. 4. T(B)=B for some B\neq 0. Please suggest which of the options are correct. It seems that 3 and -1 are the eigen values for A so (1) holds and (2) does not. - What have you tried? What problem are you running into? – copper.hat May 13 '12 at 18:52 What do you know about matrices with non-zero determinant? – Brian M. Scott May 13 '12 at 19:13 Here's a huge hint: Construct dyadic matrices from the eigenvectors of A. – copper.hat May 13 '12 at 19:35 HINT: Let v_1,v_2 be orthonormal eigenvectors of A corresponding to A's distinct eigenvalues. Now compute T(v_i v_j^T), and think about the answer. – copper.hat May 14 '12 at 0:02 @copper.hat: Thanks for the hint. But can you please illustrate it bit more. – preeti May 14 '12 at 18:10 ## 3 Answers Fixed to be about T, not about A. The trace of a square matrix equals the sum of the eigenvalues (in the algebraic closure of the ground field, if necessary; i.e., the sum of the roots of the characteristic polynomial); the determinant equals the product of the eigenvalues. This often gives a nice way of finding at least some of the eigenvalues, and in the case of 2\times 2 matrix, gives all the information required to find all the eigenvalues (since knowing a+b and ab will determine a and b). Here you have a 2\times 2 matrix, so it has two eigenvalues; their sum is 2 and their product is -3. Thus, they are 3 and -1. In particular, A is invertible, since no eigenvalue is equal to 0. Now let's consider T. Note that T is one-to-one, because if T(B)=0, then AB=0. But since A is invertible, this means that B=A^{-1}AB=A^{-1}0= 0. Thus, T is one-to-one on a finite dimensional vector space, so T is invertible. This proves that (2) is true. Now, notice that if AB=\lambda B, then A\mathbf{b}_i=\lambda\mathbf{b}_i, where \mathbf{b}_i is the ith column of A, since$$AB = A(\mathbf{b}_1\;\mathbf{b}_2) = (A\mathbf{b}_1\;A\mathbf{b}_2).$$In particular, if B is an eigenvector of T, then B\neq 0, and therefore either \mathbf{b}_1 or \mathbf{b}_2 are nonzero, so either A\mathbf{b}_1=\lambda\mathbf{b}_1 or A\mathbf{b}_2=\lambda\mathbf{b}_2 shows that \lambda is an eigenvalue of A. Conversely, if both \mathbf{b}_1 and \mathbf{b}_2 are in the eigenspace of \lambda for A, and they are not both zero, then B=(\mathbf{b}_1\;\mathbf{b}_2) is an eigenvector of T associated to \lambda. That means that the only possible eigenvalues of T are the eigenvalues of A. This proves that both (2) and (4) are false, since neither 2 nor 1 are eigenvalues of A, so they are not eigenvalues of T. The only thing left is whether T is diagonalizable. As noted above, a matrix B is an eigenvector of T associated to \lambda if and only if both columns of B lie in the eigenspace of A associated to \lambda. Since the eigenspaces of A are one-dimensional, we can select \mathbf{v}_1 and \mathbf{v}_2, eigenvectors of A associated to 3 and -1, respectively. Then B is an eigenvector of T associated to 3 if and only if B=(\alpha\mathbf{v}_1\;\beta\mathbf{v}_2) and \alpha and \beta are not both zero. Thus, we have two degrees of freedom, so the eigenspace of T associated to 3 has dimension 2. Similarly, the eigenvectors of T associated to -1 are of the form B=(\rho\mathbf{v}_2\;\sigma\mathbf{v}_2) with \rho and \sigma arbitrary but not both zero; again, the dimension is 2. Since the sum of the geometric dimensions of the eigenspaces of T is 4, which is the dimension of the vector space M_2(R), this proves that T is diagonalizable. So (1) is true. In summary, (1) and (3) are true, (2) and (4) are false. - You seem to be answering the question about A rather than T. T does not have distinct eigenvalues. – Ted May 13 '12 at 23:59 There is a difference between A and T. T has 4, not 2, eigenvalues. – copper.hat May 14 '12 at 0:00 Oops; quite so. Sorry about that. – Arturo Magidin May 14 '12 at 0:00 Hint 1: If$$ 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} $$Hint 2: \det(T)=(ad-bc)^2=\det(A)^2. Thus, replacing a and d by a-x and d-x we get that the characteristic polynomial of T is the square of the characteristic polynomial for A. That is (x^2-2x-3)^2=(x-3)^2(x+1)^2. Hint 3: Since A is diagonalizable (it has distinct eigenvalues), it has two eigenvectors which span \mathbb{R}^2: (x_1,y_1) and (x_2,y_2). (x_1,0,y_1,0), (0,x_1,0,y_1), (x_2,0,y_2,0), (0,x_2,0,y_2) are eigenvectors of T. - Note that for 2\times 2 matrices, we have the simple characteristic polynomial$$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? - 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