I'm trying to solve the following exercise in my book:
Find an orthonormal basis $\alpha$ for the vector space
$\left(\mathbb{R},\mathbb{R}^{2 \times 2},+\right)$ (with default
inner product, $ \langle A,B \rangle = Tr(A \cdot B^T $)) such that the matrix representation $L_\alpha^\alpha$
of the linear transformation $$L : \mathbb{R}^{2 \times 2} \rightarrow
\mathbb{R}^{2 \times 2} : \begin{pmatrix} x & y \\ z & t\end{pmatrix}
\mapsto \begin{pmatrix} x+y+t & x+y+z \\ y+z+t & x+z+t \end{pmatrix}$$ with respect to the basis $\alpha$ is a diagonal matrix.
I started out by transforming the default bases, as such:
$$L(e_1) = e_1 + e_2 + e_4$$
$$L(e_2) = e_1 + e_2 + e_3$$
$$L(e_3) = e_2 + e_3 + e_4$$
$$L(e_4) = e_1 + e_3 + e_4$$
With:
$$e_1 = \begin{pmatrix}1&0\\0&0\end{pmatrix}, e_2 = \begin{pmatrix}0&1\\0&0\end{pmatrix}, e_3 = \begin{pmatrix}0&0\\1&0\end{pmatrix}, e_4 = \begin{pmatrix}0&0\\0&1\end{pmatrix}$$
Can you please elaborate how to procede with such a problem. Thanks in advance!
Best Answer
I haven’t done this in quite some time, so this solution is probably unnecessary complicated:
We identify $\mathbb{R}^{2 \times 2}$ with $\mathbb{R}^4$ via $$ \mathbb{R}^{2 \times 2} \to \mathbb{R}^4, \, \begin{pmatrix} x & y \\ z & t \end{pmatrix} \mapsto (x,y,z,t)^T. $$ (So the “default basis” you used corresponds to the standard basis $(e_1, e_2, e_3, e_4)$ of $\mathbb{R}^4$.) If we understand $L$ as a linear map $\hat{L} \colon \mathbb{R}^4 \to \mathbb{R}^4$ then $\hat{L}$ is (with respect to the standard basis on both sides) given by the matrix $$ A = \begin{pmatrix} 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \end{pmatrix}. $$ Also notice that the inner product on $\mathbb{R}^{2 \times 2}$ corresponds to the standard scalar product on $\mathbb{R}^4$ because $$ \left\langle \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}, \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix} \right\rangle = a_{11} b_{11} + a_{12} b_{12} + a_{21} b_{21} + a_{22} b_{22}. $$ (This also justifies called is the default inner product.) So to find an orthonormal basis of $\mathbb{R}^{2 \times 2}$ with respect to which $L$ is diagonal is the same as finding an orthogonal basis of $\mathbb{R}^4$ with respect to which $\hat{L}$ is represented a diagonal matrix.
There are now different ways to solve this problem. We will first calculate the eigenspaces of $\hat{L}$; because $A$ is symmetric we know that $\hat{L}$ is diagonalizable. Then we will use the following fact:
So the eigenspaces of different eigenvalues are orthogonal to each other. Therefore we can compute for each eigenspace an orthonormal basis and them put them together to get one of $\mathbb{R}^4$; then each basis vectors will in particular be an eigenvectors $\hat{L}$.
By some lengthy calculation it can be shown that the characteristic polynomial of $A$ is given by $$ \chi_A(t) = t^4 - 4 t^3 + 2 t^2 + 4t - 3. $$ It is easy to guess the roots $1$ and $-1$, so we can factor $\chi_A$ and get $$ \chi_A(t) = (t-1)^2 (t+1) (t-3). $$ The eigenspaces can now be calculated as usual, and we find that $$ E_1 = \langle (0,-1,0,1)^T, (-1,0,1,0)^T \rangle, \; E_{-1} = \langle (-1,1,-1,1)^T \rangle, \; E_3 = \langle (1,1,1,1)^T \rangle, $$ where $E_\lambda$ denotes the eigenspace with respect to the eigenspace $\lambda$.
Next we need to find orthonormal basis for each eigenspace. We can always do this by picking some basis and then using Gram–Schmidt. But here we are pretty lucky:
We know the basis $((0,-1,0,1)^T, (-1,0,1,0)^T)$ of $E_1$. Because both basis vectors are already orthogonal to each other we only need to normalize them. So we get $b_1 = \frac{1}{\sqrt{2}}(0,-1,0,1)^T$ and $b_2 = \frac{1}{\sqrt{2}}(-1,0,1,0)^T$.
In the case of $E_{-1}$ and $E_3$ we are even luckier, as they are both one-dimensional. So here too we only need to normalize and thus get $b_3 = \frac{1}{2} (-1,1,-1,1)^T$ and $b_4 = \frac{1}{2}(1,1,1,1)^T$.
Putting these together we have now found a basis $(b_1, b_2, b_3, b_4)$ of $\mathbb{R}^4$ given by $$ b_1 = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 \\ -1 \\ 0 \\ 1 \end{pmatrix}, \; b_2 = \frac{1}{\sqrt{2}} \begin{pmatrix} -1 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \; b_3 = \frac{1}{2} \begin{pmatrix} -1 \\ 1 \\ -1 \\ 1 \end{pmatrix}, \; b_4 = \frac{1}{2} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}, $$ which is orthonormal and cosists of eigenvectors of $\hat{L}$. The corresponding $2 \times 2$ matrices are \begin{align*} B_1 &= \frac{1}{\sqrt{2}} \begin{pmatrix} 0 & -1 \\ 0 & 1 \end{pmatrix}, & B_2 &= \frac{1}{\sqrt{2}} \begin{pmatrix} -1 & 0 \\ 1 & 0 \end{pmatrix}, \\ B_3 &= \frac{1}{2} \begin{pmatrix} -1 & 1 \\ -1 & 1 \end{pmatrix}, & B_4 &= \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}. \end{align*}