# Find $2\times 2$ symmetric matrix $A$ given two eigenvalues and one eigenvector

I am having trouble finding the symmetric matrix $$A$$ given eigenvalues $$1$$ and $$4$$ and eigenvector $$(1, 1)$$ corresponding to eigenvalue $$1$$.

I feel like I'd have to use the equation $$A=PD(P^{-1})$$, but I'm having trouble finding the matrix $$P$$ if I can't find the second eigenvector. Any help is appreciated, thanks!

• Different eigenspaces of a symmetric matrix are orthogonal to each other. Dec 7 '18 at 21:54
• ...thus $[1,-1]^T$ could be this vector (attached to eigenvalue 4)... Dec 7 '18 at 22:04

To offer a slightly different perspective, due to the Spectral Theorem you have $$A=P_1+4P_2,$$ where $$P_1$$ is the orthogonal projection onto the span of $$(1,1)^t$$, and $$P_2$$ is orthogonal to $$P_1$$; that is, $$P_2=I-P_1$$. Thus $$P_1=\tfrac12\,\begin{bmatrix} 1&1\end{bmatrix}\begin{bmatrix} 1\\1\end{bmatrix}=\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}.$$ And then $$A=\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}+4\begin{bmatrix}1/2&-1/2\\-1/2&1/2\end{bmatrix}=\begin{bmatrix}5/2&-3/2\\-3/2&5/2 \end{bmatrix}$$

Let :

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

Since the eigenvalues are $$\lambda =1$$ and $$\lambda = 4$$, it must be

$$\det(A-\lambda I) = \begin{vmatrix} a - \lambda & b \\ c & d - \lambda\end{vmatrix} = (a-\lambda)(d-\lambda)-bc$$

such that $$\lambda = 1$$ and $$\lambda = 4$$ are solutions to the equation :

$$(a-\lambda)(d-\lambda)-bc=0$$

Since $$(1,1)^\mathbf{T}$$ is an eigenvector of $$A$$ for $$\lambda = 1$$, it is :

$$(A- I)(1,1)^\mathbf{T} = 0 \Rightarrow \begin{pmatrix} a - 1 & b \\ c & d - 1\end{pmatrix}\begin{pmatrix}1 \\ 1 \end{pmatrix} =\begin{pmatrix}0 \\ 0 \end{pmatrix}$$

$$\Leftrightarrow$$

$$\begin{cases} a-1 + b = 0 \\c + d-1 = 0\end{cases}$$

Can you now find $$a,b,c$$ and $$d$$ ?

• Also use the other two conditions: $\lambda_1 \lambda_2=4=ad-bc$ and $\lambda_1+\lambda_2=4+1=trace (A)=a+d$.. i.e., $a+d=5, \ ad-bc=4$.
– Why
Dec 7 '18 at 22:10