Show $A$ is Hermitian and find the orthonormal basis for $V$ in which $A$ is diagonalizable.

Question

Let $\{e_1,e_2,e_4\}$ be an orthonormal basis for a complex unitary space $V$. Let's define the vectors: $f_j=e_j-\frac14\sum\limits_{i=1}^4e_i, j\in\{1,2,3,4\}$. Let $A\in\mathcal L(V), Ax:=\sum\limits_{j=1}^4\langle x,f_j\rangle f_j$.

Show $A$ is Hermitian and find the orthonormal basis for $V$ in which $A$ is diagonalizable.

Note: typo corrected.

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My attempt:

Let's compute $f_1,f_2,f_3,f_4$ first.

$\begin{aligned}f_j=e_j-\frac14\sum\limits_{i=1}^4, e_i\implies&f_1=\frac34e_1-\frac14(e_2+e_3+e_4)\\&f_2=\frac34e_2-\frac14(e_1+e_3+e_4)\\&f_3=\frac34e_3-\frac14(e_1+e_2+e_4)\\&f_4=\frac34e_4-\frac14(e_1+e_2+e_3)\end{aligned}$

$\begin{aligned}Ae_i&=\sum\limits_{j=1}^4\langle e_i,f_j\rangle f_j\implies Ae_1=\left\langle e_1,\frac34e_1-\frac14(e_2+e_3+e_4)\right\rangle f_1+\left\langle e_1,\frac34e_2-\frac14(e_1+e_3+e_4)\right\rangle f_2+\left\langle e_1,\frac34e_3-\frac14(e_1+e_2+e_4)\right\rangle f_3+\left\langle e_1,\frac34e_4-\frac14(e_1+e_2+e_3)\right\rangle f_4=\frac34f_1-\frac14(f_2+f_3+f_4)\end{aligned}$

$\ Ae_2=\frac34f_2-\frac14(f_1+f_3+f_4)\\Ae_3=\frac34f_3-\frac14(f_1+f_2+f_4)\\Ae_4=\frac34f_4-\frac14(f_1+f_2+f_3)$

Then, $$[A]_e^f=\begin{bmatrix}\frac34&-\frac14&-\frac14&-\frac14\\-\frac14&\frac34&-\frac14&-\frac14\\-\frac14&-\frac14&\frac34&-\frac14\\-\frac14&-\frac14&-\frac14&\frac34\end{bmatrix}$$

About the matrix representation of $A\in\mathcal L(V)$:

$A\in M_n(\Bbb R)\ \&\ A=A^\tau\ \implies A=A^*\iff A\ \text{is normal}\implies A\text{ is diagonalizable in some orthonormal basis}$ $\{a_1,a_2,a_3,a_4\}$

Let's find the eigenvalues and the corresponding eigenspaces using the formula derived here. According to the notation I used in the thread, $a_j=\frac34-\lambda\ \forall j\in\{1,2,3,4\}$ and $x=-\frac14$.

$$\det(A-\lambda I)=\begin{vmatrix}\frac34-\lambda&-\frac14&-\frac14&-\frac14\\-\frac14&\frac34-\lambda&-\frac14&-\frac14\\-\frac14&-\frac14&\frac34-\lambda&-\frac14\\-\frac14&-\frac14&-\frac14&\frac34-\lambda\end{vmatrix}=\left(\frac34-\lambda+\frac14\right)^4\left(1-\frac14\cdot 4\cdot\frac1{\frac34-\lambda+\frac14}\right)=-\lambda(1-\lambda)^3=\lambda(\lambda-1)(1-\lambda)^2\implies\sigma(A)=\{0,1\}$$ Let's use the fact $\Omega$ is the orthonormal complement of the row-space (explanation for the indeces orther I used) since $\boxed{E_A(0)\oplus E_A(1)=V}$:

Now, $E_A(0)=\ker(A)$: $$\begin{bmatrix}\frac34&-\frac14&-\frac14&-\frac14\\-\frac14&\frac34&-\frac14&-\frac14\\-\frac14&-\frac14&\frac34&-\frac14\\-\frac14&-\frac14&-\frac14&\frac34\end{bmatrix}\sim\begin{bmatrix}1&1&1&-3\\1&1&-3&1\\1&-3&1&1\\-3&1&1&1\end{bmatrix}\sim\begin{bmatrix}1&1&1&-3\\0&0&-4&4\\0&-4&0&4\\0&4&4&-8\end{bmatrix}\sim\begin{bmatrix}1&1&1&-3\\0&0&-1&1\\0&-1&0&1\\0&0&0&0\end{bmatrix}\sim\begin{bmatrix}1&0&0&-1\\0&-1&0&1\\0&0&-1&1\\0&0&0&0\end{bmatrix}$$

$$\implies E_A(0)=\operatorname{span}\left\{\underbrace{\begin{bmatrix}1\\1\\1\\1\end{bmatrix}}_{v_4}\right\}$$

$E_A(1)=\ker(A-I)$: $$\begin{bmatrix}-\frac14&-\frac14&-\frac14&-\frac14\\-\frac14&-\frac14&-\frac14&-\frac14\\-\frac14&-\frac14&-\frac14&-\frac14\\-\frac14&-\frac14&-\frac14&-\frac14\end{bmatrix}\sim\begin{bmatrix}1&1&1&1\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}$$

$$\implies E_A(1)=\operatorname{span}\left\{\underbrace{\begin{bmatrix}1\\0\\0\\-1\end{bmatrix}}_{v_1},\underbrace{\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}}_{v_2},\underbrace{\begin{bmatrix}0\\0\\-1\\1\end{bmatrix}}_{v_3}\right\}$$

Let's apply Gramm-Schmidt to the obtained basis for $V$: $$a_1=\frac1{\|v_1\|}v_1=\frac1{\sqrt{2}}\begin{bmatrix}1\\0\\0\\-1\end{bmatrix}=b_1$$ $$\begin{aligned}b_2&=v_2-\langle v_2,a_1\rangle a_1\\&=\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}-\frac1{\sqrt{2}}\left\langle\begin{bmatrix}0\\-1\\0\\1\end{bmatrix},\begin{bmatrix}1\\0\\0\\-1\end{bmatrix}\right\rangle\frac1{\sqrt{2}}\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}\\&=\frac32\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}\end{aligned}$$ $$a_2=\frac1{\|b_2\|}b_2=\frac1{\sqrt{2}}\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}$$ $$\begin{aligned}b_3&=v_3-\langle v_3,a_1\rangle a_1-\langle v_3,a_2\rangle a_2\\&=\begin{bmatrix}0\\0\\-1\\1\end{bmatrix}-\frac1{\sqrt{2}}\left\langle\begin{bmatrix}0\\0\\-1\\1\end{bmatrix},\begin{bmatrix}1\\0\\0\\-1\end{bmatrix}\right\rangle\frac1{\sqrt{2}}\begin{bmatrix}1\\0\\0\\-1\end{bmatrix}-\frac1{\sqrt{2}}\left\langle\begin{bmatrix}0\\0\\-1\\1\end{bmatrix},\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}\right\rangle\frac1{\sqrt{2}}\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}\\&=\begin{bmatrix}0+\frac12\\0+\frac12\\-1\\1-\frac12-\frac12\end{bmatrix}=\begin{bmatrix}\frac12\\\frac12\\-1\\0\end{bmatrix}\end{aligned}$$ $$a_3=\frac1{\|b_3\|}b_3=\frac{\sqrt{6}}3\begin{bmatrix}\frac12\\\frac12\\-1\\0\end{bmatrix}$$ $$\begin{aligned}b_4&=v_4-\langle v_4,a_1\rangle a_1-\langle v_4,a_2\rangle a_2-\langle v_4,a_3\rangle a_3\\&=\begin{bmatrix}1\\1\\1\\1\end{bmatrix}-\frac1{\sqrt{2}}\left\langle\begin{bmatrix}1\\1\\1\\1\end{bmatrix},\begin{bmatrix}1\\0\\0\\-1\end{bmatrix}\right\rangle\frac1{\sqrt{2}}\begin{bmatrix}1\\0\\0\\-1\end{bmatrix}-\frac1{\sqrt{2}}\left\langle\begin{bmatrix}1\\1\\1\\1\end{bmatrix},\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}\right\rangle\frac1{\sqrt{2}}\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}-\frac{\sqrt{6}}3\left\langle\begin{bmatrix}1\\1\\1\\1\end{bmatrix},\begin{bmatrix}\frac12\\\frac12\\-1\\0\end{bmatrix}\right\rangle\frac{\sqrt{6}}3\begin{bmatrix}\frac12\\\frac12\\-1\\0\end{bmatrix}\\&=\begin{bmatrix}1\\1\\1\\1\end{bmatrix}\ \underline{\text{we can skip this step}}\end{aligned}$$ $$a_4=\frac1{\|b_4\|}b_4=\frac12\begin{bmatrix}1\\1\\1\\1\end{bmatrix}$$

Therefore, a Hermitian operator $A$ is diagonalizable in the orthonormal basis: $$\{a_1,a_2,a_3,a_4\}=\left\{\frac1{\sqrt{2}}\begin{bmatrix}1\\0\\0\\-1\end{bmatrix},\frac1{\sqrt{2}}\begin{bmatrix}0\\-1\\0\\1\end{bmatrix},\frac{\sqrt{6}}3\begin{bmatrix}\frac12\\\frac12\\-1\\0\end{bmatrix},\frac12\begin{bmatrix}1\\1\\1\\1\end{bmatrix}\right\}$$

May I ask if this is correct? If so, how could I improve my approach?

Thank you in advance!

Answer 1

Your computations are correct. I'll show a more conceptual approach below.

For the selfadjoint part, one can easily show that $(\langle\cdot,x\rangle\,y)^*=\langle\cdot,y\rangle\,x$. So $A$ is selfadjoint.

Let $e=\tfrac14\,\sum_je_j$. Then \begin{align} Ax&=\sum_j\langle x,e_j-e\rangle,(e_j-e)=\sum_j\langle x,e_j\rangle\,e_j+4\langle x,e\rangle\,e-\sum_j\langle x,e_j\rangle \,e-\sum_j\langle x,e\rangle\,e_j\\[0.3cm] &=x+4\langle x,e\rangle\,e-\langle x,4e\rangle e-\langle x,e\rangle\,4e\\[0.3cm] &=x-4\langle x,e\rangle\,e=x-\langle x,e'\rangle\,e', \end{align} where $e'=2e$. This gives us $\|e'\|=1$, so the rank-one operator $P:x\longmapsto \langle x,e'\rangle\,e'$ is a rank-one projection.

Thus $A=I-P$ for a rank-one projection $P$, and $A$ is then a rank-three projection. This already gives us that $A$ is selfadjoint and that its eigenvalues are $\{1,1,1,0\}$, but we don't really need this here.

Since the identity is already diagonal in any basis, we only need to diagonalize $P$. We achieve this by constructing a basis $\{e',g_2,g_3,g_4\}$. We may then take for instance \begin{align} e'&=\frac12\,(e_1+e_2+e_3+e_4),\ \ \ \ g_2=\frac12\,(-e_1-e_2+e_3+e_4),\\[0.3cm] g_3&=\frac12\,(e_1-e_2-e_3+e_4),\ \ \ \ g_4=\frac12\,(-e_1+e_2-e_3+e_4),\\[0.3cm] \end{align} which gives us $$ Ae'=0,\ \ Ag_j=g_j,\ \ \ j=2,3,4. $$ There is a lot of freedom to choose $g_2,g_3,g_4$, as any orthonormal basis of the 3-dimensional orthogonal complement of $\{e\}$ will do.