Find a $2 \times 2$ unitary matrix with eigenvalues $1, -1$ and the eigenvectors $$\begin{pmatrix}1\\ 0\end{pmatrix},\:\begin{pmatrix}0\\ 1\end{pmatrix}$$

My attempt:

Let $$A = \begin{pmatrix} a_1 & a_2\\ a_3 & a_4\end{pmatrix}$$ so that $$\det\begin{pmatrix}a_1-\lambda &a_2\\ a_3&a_4-\lambda \end{pmatrix} = 0$$ but it's look like a bit difficult to continue, maybe there is a simple way?

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    $\begingroup$ Observe that the matrix is determined by how it acts on a basis. You're given how it acts on the standard basis. Does this help you? $\endgroup$ – Johnny El Curvas Apr 23 at 9:36
  • $\begingroup$ What are the difference between the basis and the standard basis? $\endgroup$ – Xavi Apr 23 at 9:36
  • $\begingroup$ There are many bases for the same space (in this case $\mathbb{C}^{2}$. By standard basis I mean the one formed by $(1,0)$ and $(0,1)$. $\endgroup$ – Johnny El Curvas Apr 23 at 9:37
  • $\begingroup$ In the question, it is given that $A(1,0)=(1,0)$ and $A(0,1)=(0,-1)$. Write these equations in terms of the coefficients of $A$ and see what you can get. $\endgroup$ – Johnny El Curvas Apr 23 at 9:39
  • $\begingroup$ @JohnnyElCurvas But how you find it in an easier way? $\endgroup$ – Xavi Apr 23 at 9:39

Let $A\in \mathbb{C}^{2\times 2}$ be a unitary matrix such that its eigenvalues are $1$ and $-1$ with eigenvectors $(1,0)$ and $(0,1)$. This means (by definition), that $A(1,0)^{T}=(1,0)^{T}$ and $A(0,1)^{T}=-(0,1)^{T}$ (here the $T$ just means transposing). Let's write

$$ A=\begin{pmatrix} a & b \\ c & d \end{pmatrix} $$

The equations above imply that

$$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} =\begin{pmatrix} a \\ c \end{pmatrix} =\begin{pmatrix} 1 \\ 0 \end{pmatrix} $$ and $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} =\begin{pmatrix} b \\ d \end{pmatrix} =\begin{pmatrix} 0 \\ -1 \end{pmatrix} $$

Now, componentwise, you get $a=1$, $c=0$, $b=0$, $d=1$, so that

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

Now, this matrix has eigenvalues $1$ and $-1$, and eigenvectors $(1,0)$ and $(0,1)$, as desired. It just remains to check that indeed $A$ is unitary. For example, you can argue that this is the case because the columns of $A$ form an orthonormal basis for $\mathbb{C}^{2}$, or you can simply compute $A^{*}A$ and check that it is the identity matrix.

Hope this helps!


Generally, if you know that a linear operator$\phi$ is diagonalisable with eigenvalues $\lambda_1,\ldots,\lambda_n$ with respect to some (ordered) basis$~\mathcal B$ of eigenvectors, this means precisely that the matrix $\operatorname{Mat}_{\mathcal B}(\phi)$ of the operator with respect to that basis is diagonal, with diagonal entries $\lambda_1,\ldots,\lambda_n$. Generally, if an operator that acts on $K^n$ (where $K$ is your field, probably $K=\Bbb C$ when you are talking about unitary matrices) and is given by is matrix $A$, then this means that $A$ is the matrix of $\phi$ with respect to the standard basis of $K^n$. If one is in both these situations at once (so $\phi$ is a diagonalisable linear operator acting on $K^n$ given by its matrix $A$), and the basis$~\mathcal B$ of eigenvectors differs from the standard basis, then the diagonal matrix $\operatorname{Mat}_{\mathcal B}(\phi)$ differs from $A$ and is related to it by a change of basis operation. Here however you are lucky in that it is given that the basis of eigenvectors is equal to the standard basis; then no conversion is necessary, and $A$ is simply the diagonal matrix with the eigenvalues, in order, as diagonal entries.

As you should know, a change of basis, if necessary, would be performed by forming the matrix $P$ whose columns describe the vectors of $\mathcal B$ (with respect to the standard basis), which is invertible (since $\mathcal B$ is a basis), and if $D$ is the diagonal matrix mentioned, then one has the relation $A=PDP^{-1}$ (and $D=P^{-1}AP$). But with $P=I$ this of course gives $A=D$.


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