Prove that there is a base of $\mathbb R^4$ made of eigenvectors of matrix $A$

Matrix of linear operator $\mathcal A$:$\mathbb R^4$ $\rightarrow$ $\mathbb R^4$ is $$A= \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & -1 & -1\\ 1 & -1 & 1 & -1\\ 1 & -1 & -1 & -1\\ \end{bmatrix}$$ Prove that there is a base of $\mathbb R^4$ made of eigenvectors of matrix $A$. Using the new base, find matrix of that operator.

I hope I translated all correctly.

This is what I have done so far.

1. I found characteristic polynomial of matrix $A$, so I can get eigenvalues and thus find eigenvectors. My characteristic polynomial is $$p_A(\lambda)=\lambda^4-2\lambda^3-6\lambda^2+16\lambda-8$$
2. My eigenvalues are $$\lambda_1=\lambda_2=2$$ $$\lambda_3=-1-\sqrt3$$ $$\lambda_4=-1+\sqrt3$$
3. After that I calculated my eigenvectors. This is where I need help understanding. Eigenvectors that belong to different eigenvalues are linearly independent so then they can make a base. In this case, I have two equal eigenvalues. But, when I calculate: $$A\overrightarrow v=\lambda_1 \overrightarrow v$$ where $\overrightarrow v=(x_1,x_2,x_3,x_4)$ is eigenvector for eigenvalue 2 I get this form (final):$$[A-\lambda_1I]= \begin{bmatrix} -1 & 1 & 1 & 1 \\ 0 & 0 & 0 & -2\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{bmatrix}$$

So, my vector $$\overrightarrow v= \begin{bmatrix} x_2+x_3 \\ x_2\\ x_3\\ 0\\ \end{bmatrix}=x_2\begin{bmatrix} 1 \\ 1\\ 0\\ 0\\ \end{bmatrix}+x_3\begin{bmatrix} 1 \\ 0\\ 1\\ 0\\ \end{bmatrix}$$

So, I am not even sure how to ask this question. Even if eigenvalues where the same, I did get one vector that is actually a linear combination of two linearly independent vectors? Is this observation correct?

After that I calculated eigenvectors for remaining eigenvalues. These were results:

$\overrightarrow v_3 = x'_4\begin{bmatrix} -\sqrt3 \\ \sqrt3\\ \sqrt3\\ 1\\ \end{bmatrix}$ where $\overrightarrow v_3=(x'_1,x'_2,x'_3,x'_4)$ for $\lambda_3=-1-\sqrt3$

$\overrightarrow v_4 = x''_4\begin{bmatrix} \sqrt3 \\ -\sqrt3\\ -\sqrt3\\ 1\\ \end{bmatrix}$ where $\overrightarrow v_4=(x''_1,x''_2,x''_3,x''_4)$ for $\lambda_4=-1+\sqrt3$

So, in this case, is my base:

$$B= \begin{bmatrix} 1 & 1 & -\sqrt3 & \sqrt3 \\ 1 & 0 & \sqrt3 & -\sqrt3\\ 0 & 1 & \sqrt3 & -\sqrt3\\ 0 & 0 & 1 & 1\\ \end{bmatrix}$$?

and would new matrix of operator $\mathcal A$ be $B^{-1}AB$?

I also have one more question: Is there some shorter way in finding these results? I am not lazy to do these calculations, but it is easy to make a mistake when time is short. Could I conclude something by looking at matrix $A$ to help me find eigenvalues and eigenvectors faster?

Thank you all in advance.

• You actually proved the eigenspace $E_2$ has dimension $2$, so the geometric multiplicity of the eigenvalue $2$ is equal to its algebraic multiplicity, and the matrix is diagonalisable. I don't think there is any shorter way. – Bernard May 31 '16 at 16:42
• Your questions is perfectly valid as it is, but I think there's a tiny chance you copied it wrongly, and that the last diagonal entry of $A$ is actually $1$ (not $-1$) — for purely aesthetic reasons. Then the columns of $A$ form an orthogonal (but not orthonormal) basis of $\mathbb R^4$. Again, your question as it is at present is perfectly alright! – M. Vinay Jun 1 '16 at 2:07
• @M.Vinay Nice to know my question makes sense. Oh, and it is $-1$ indeed but I see your point. – Asleen Jun 1 '16 at 14:46
• @Asleen Ah, okay. But it would make an interesting question if it were $+1$ :) – M. Vinay Jun 1 '16 at 14:49

Note that your matrix $A$ is symmetric and hence diagonalizable. You don't even need to find the eigenvalues of $A$ to conclude that there exists a basis of eigenvectors for $A$. I don't see any calculation-free way to find the eigenvalues of $A$ but once you find them, you don't need to know the eigenvectors in order to know how the operator will look with respect to a basis of eigenvectors. If the eigenvectors are $v_1, \dots, v_4$ with $Av_i = \lambda_i v_i$ then with respect to $(v_1, \dots, v_4)$ the operator will be $\operatorname{diag}(\lambda_1, \dots, \lambda_4)$.

If you are not asked explicitly to find a basis of eigenvectors for $A$, you can skip 3 entirely and say that $A$ can be represented as $\operatorname{diag}(2,2,-1-\sqrt{3},-1+\sqrt{3})$ (or by any matrix that is obtained by permuting the rows).

Last comment - the trace of your matrix is 2 and this should be the sum of the eigenvalues $\lambda_1 + \dots + \lambda_4$. This can be used for "sanity check" after calculating the eigenvalues to make sure you haven't done a computation error (this doesn't guarantee that you haven't made a mistake but provides some evidence for it).

• Thank you very much. This does help a lot. Is there any chance you could tell me am I right about the last question? Would new matrix of operator be matrix $B^{-1}AB$? Thank you. – Asleen Jun 1 '16 at 14:35
• Yes, the new matrix will indeed be $B^{-1}AB$. Note that you don't have to calculate $B^{-1}$ and perform the multiplication explicitly since you know that if you haven't made a mistake, the result will be $\operatorname{diag}(2,2,-1-\sqrt{3},-1+\sqrt{3})$. – levap Jun 1 '16 at 18:59

Carl Meyer

$\S$ 7.2, eqn 7.2.5, p 512

Diagonalizability and Multiplicities

The matrix $\mathbf{A}\in\mathcal{C}^{n\times n}$ is diagonalizable iff $$geometric\ multiplicity _{\mathbf{A}} \left( \lambda \right) = algebraic\ multiplicity _{\mathbf{A}} \left( \lambda \right)$$ for each $\lambda\in\sigma \left( \mathbf{A} \right)$. That is, iff every eigenvalue is semisimple.

Application

You have identified the eigenvalues that their algebraic multiplicities. The issue is to quantify the geometric multiplicity of the eigenvalue $\lambda = 2$.

The geometric multiplicity $$geometric\ multiplicity _{\mathbf{A}} \left( 2 \right) = \dim N \left( \mathbf{A} - 2 \mathbf{I}_{\,4} \right)$$ $$\mathbf{A} - 2 \mathbf{I}_{\,4} = \left[ \begin{array}{rrrr} -1 & 1 & 1 & 1 \\ 1 & -1 & -1 & -1 \\ 1 & -1 & -1 & -1 \\ 1 & -1 & -1 & -3 \\ \end{array} \right]$$ The row reduction process is immediate and leaves $$\left[ \begin{array}{rrrr} 1 & -1 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right].$$ The rank of this matrix is 2; therefore the geometric multiplicity is 2. Therefore $$geometric\ multiplicity _{\mathbf{A}} \left( 2 \right) = algebraic\ multiplicity _{\mathbf{A}} \left( 2 \right)$$ and $\mathbf{A}$ is diagonalizable.