# Find a unitary matrix given eigenvectors and eigenvalues

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?

• 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? – Johnny El Curvas Apr 23 at 9:36
• What are the difference between the basis and the standard basis? – Xavi Apr 23 at 9:36
• 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 $(0,1)$. – Johnny El Curvas Apr 23 at 9:37
• 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. – Johnny El Curvas Apr 23 at 9:39
• @JohnnyElCurvas But how you find it in an easier way? – Xavi Apr 23 at 9:39

## 2 Answers

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 $$A=D$$.