Finding the sec. Eigenvector, when knowing the first Eigenvector and Eigenvalue

Question

So here is my problem,

Let A be a symmetric Matrix (2x2) with EV=(-2, 3) and the Eigenvalue being -5.

Find the 2. Eigenvector to the second Eigenvalue.

The only info i can think of is that the EV of a symmetric matrix have to be orthogonal, so the dot product has to be 0.

So -2x1 +3x2 =0 I then can fix x1=a for example and then calculate, x2=2/3

$$ (1, 2/3)^t *a, $$ for all a Elements of R without 0

Is this the correct answer it seems a bit too simple....

Answer 1

You know that $\begin{pmatrix}-2\\3\end{pmatrix}$ is an eigenvector relative to $-5$. You correctly argue that the eigenvectors relative to the other eigenvalues are orthogonal to this one, so we find $-2x_1+3x_2=0$. Any such vector is good, so we can take $\begin{pmatrix}3\\2\end{pmatrix}$.

You are correct.

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The spectral theorem says that $$ A=-5vv^T+\lambda ww^T $$ where $v$ and $w$ are the normalization of the two eigenvectors. There is no way to determine the second eigenvalue from the given data. You have $$ A=-\frac{5}{13}\begin{pmatrix} 4 & -6 \\ -6 & 9\end{pmatrix} +\frac{\lambda}{13}\begin{pmatrix} 9 & 6 \\ 6 & 4\end{pmatrix} $$

Answer 2

If $$M=\begin{pmatrix} a & b \\ b & c \end{pmatrix}.$$

Then $\lambda+ \lambda ' = a+c$ and $\lambda \lambda ' = ac-b^2$ and $$-2a+3b =10$$ $$-2b+3c =-15$$

Let $b=6t$ and $\lambda ' = -5$ then $$a= 9t-5$$ $$c= 4t-5$$

so we have $\lambda = 13t-5$. Now we see that your matrix is not uniqely determined: $$M=\begin{pmatrix} 9t-5 & 6t \\ 6t & 4t-5 \end{pmatrix}.$$

Answer 3

Your solution and reasoning are correct. Every $2\times2$ symmetric matrix has a pair of orthogonal eigenvectors. Thus, having one of the two eigenvectors specifies the other, as in two dimensions there are only two orthogonal directions.

If you want to see this algebraically, it follows from \begin{multline} \left[\begin{matrix} a & b \\ b & c\end{matrix}\right]\left[\begin{matrix} x \\ y\end{matrix}\right] = \lambda_1 \left[\begin{matrix} x \\ y\end{matrix}\right]\Longrightarrow\begin{matrix}a x + b y = \lambda_1 x \\ b x + c y = \lambda_1 y\end{matrix}\Longrightarrow\begin{matrix} (c-\lambda_1)y + b x =0 \\ -b y +(\lambda_1 - a)x = 0\end{matrix} \\ \Longrightarrow\begin{matrix} -(a - a - c + \lambda_1)y + b x =0 \\ -b y +(c - c + \lambda_1 - a)x = 0\end{matrix}\Longrightarrow\begin{matrix} -ay + b x = -(a+c-\lambda_1)y \\ -b y +c x = (a + c -\lambda_1) x \end{matrix}\\\Longrightarrow\left[\begin{matrix} a & b \\ b & c\end{matrix}\right]\left[\begin{matrix} -y \\ x\end{matrix}\right]=(a + c - \lambda_1) \left[\begin{matrix} -y \\ x\end{matrix}\right]. \end{multline}