Finding if $f$ (a linear transformation) is diagonalizable

Question

Let $f:{\mathcal M}_2(\mathbb{R})\to{\mathcal M}_2(\mathbb{R}), f(X)= X- X^t.$ Find if $f$ is diagonalisable.

We take the canonic base in ${\mathcal M}_2(\mathbb{R})$, $B=\{{\begin{pmatrix}1&0\\0&0\end{pmatrix},\begin{pmatrix}0&1\\0&0\end{pmatrix},\begin{pmatrix}0&0\\1&0\end{pmatrix},\begin{pmatrix}0&0\\0&1\end{pmatrix}}\}$. Let's call them in order $E_1,E_2,E_3,E_4.$

Now to find $M_f$ in $B$. We have to do the following:

$$f(E_1)=\lambda_{1,1}E_1+...+\lambda_{1,4}E_4$$ $$...$$ $$f(E_4)=\lambda_{4,1}E_1+...+\lambda_{4,4}E_4$$

And $M_f$ is all those $\lambda$'s on columns. Well then $M_f$ will be:

$$M_f=\begin{pmatrix}0&0&0&0\\0&1&-1&0\\0&-1&1&0\\0&0&0&0\end{pmatrix}$$

So then to find it's eigenvalues we have to find it's characteristic polynomial, to do that, we have that:

$$P_{M_f}(x)=det(M_f-xI_4)=x^3(x-2).$$

So we have eigenvalues: $\lambda_1=0$ with the algebraic multiplicity $a(\lambda_1)=3$ and $\lambda_2=2$ with $a(\lambda_2)=1.$ We now want to find $V_{\lambda_{1,2}}.$ To do that, don't we have to $(M_f-\lambda_1I_4)\begin{pmatrix}x\\y\\z\\t\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\end{pmatrix}$?

Because that gives me $V_{\lambda_1}=\{\alpha\begin{pmatrix}0&1\\1&0\end{pmatrix}|\alpha\in\mathbb{R}\}.$

And that's not correct, what am I doing wrong?

Answer 1

You are not doing anything wrong yet. Next, you need to find the other eigenspace and if its dimension is $3$, then the matrix is readily diagonalizable. But if not, then the best you can do is you can get a Jordan normal form.

I think you are being confused about the eigenvectors. Note that, once you represent a linear transformation as a matrix with a fixed basis, then the eigenvectors you are finding are of that matrix. As such, they are elements of $\mathbb{R^n}$, if the original linear transformation is acting on a vector space with dimension $n.$

Answer 2

Note that $X-X^t=0$ if and only of $X$ is symmetric. It easy to see that the subspace $\mathcal{S}_2(\Bbb R)$ of symmetric matrices in $\mathcal{M}_2(\Bbb R)$ is generated by $$ S_1=\begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix},\quad S_2=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix},\quad S_3=\begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}$$

Hint:

Write any symmetric matrix as $$\begin{pmatrix} a & b\\ b & c \end{pmatrix}.$$ Any symmetric matrix depend on $3 $parameters $a,b,c$; set $a=1, b=c=0$ to find $S_1$. Similarly $b=1, a=c=0$ for $S_2$ and $a=b=0,c=1$ for $S_3$. To check that any symmetric matrix is linear combination of $S_1,S_2,S_3$ just notice that $$\begin{pmatrix} a & b\\ b & c \end{pmatrix}=a S_1+ b S_2+ c S_3.$$

Complete $\{S_1,S_2,S_3\}$ to a basis of $\mathcal{M}_2(\Bbb R)$ by adding the matrix $$A=\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}$$ It easy to verify that $\{S_1,S_2,S_3,A\}$ is a basis for $\mathcal{M}_2(\Bbb R)$.

Hint:

For any matrix $X$ consider $$ S_X=\frac12 (X+X^t) \quad A_X=\frac12( X-X^t).$$ By definition $S_X$ is symmetric and $A_X$ is antisymmetric. Now use the first hint to write $S_X$ as a linear combination of $\{S_1,S_2,S_3\}$. $A_X$ is clearly a real multiple of $A$.

Now use this basis to write $\quad f:\mathcal{M}_2(\Bbb R)\longrightarrow\mathcal{M}_2(\Bbb R)$, $\quad f(X)=X-X^t$. What you find s the following matrix

$$\begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 2\\ \end{pmatrix}$$

which is clearly diagonal, hence $f$ is diagonalizable.

Answer 3

As other people have stated, you only needed to find eigenspaces, which are the subspace of symmetric matrices (for $\lambda =0$) and the subspace of antisymmetric matrices (for $\lambda=2$). Since these subspaces have dimensions $3$ and $1$, and $3+1=4=2^2$, $f$ is therefore diagonalizable. There is an alternative way. You can find the minimal polynomial of $f$ directly from its definition.

Note that \begin{align}f^2(X)&=f(X-X^t)=(X-X^t)-(X-X^t)^t\\&=(X-X^t)-(X^t-X)=2(X-X^t)=2f(X).\end{align} Thus, $f^2-2f=0$. That is, the minimal polynomial of $f$ divides $x^2-2x=x(x-2)$. Since $f(I)=0$ and $f(A)=2A$, where $I$ is the identity matrix and $A$ is the antisymmetric matrix in InsideOut's solution, we get that both $0$ and $2$ are eigenvalues of $f$. Thus, the minimal polynomial of $f$ is precisely $x(x-2)$. Since the roots of $x(x-2)$ are simple, $f$ is diagonalizable.

In fact, you don't even need to know the eigenvalues of $f$. If a linear transformation $T:V\to V$ on a vector space over a field $K$ satisfies $p(T)=0$ for some polynomial $p(x)$ whose roots are all in $K$ and are simple, then $T$ is diagonalizable.