Prove u is an eigenvector of matrix $A$ (if $a+b=c+d$) and find the eigenvalues of $A$

Question

a. Prove $u=\begin{bmatrix}1\\1\end{bmatrix}$ is an eigenvector of matrix $A$, if $a+b=c+d:$ \begin{bmatrix}a&b\\c&d\end{bmatrix} b. Find the eigenvalues of matrix $A$

I was able to prove a. and find $\lambda_1=a+b$.

Answer 1

For (a) just calculate $Au$ and show that $Au=(a+b) u$ if $a+b=c+d$.

For (b) use the fact that the sum of the eigenvalues of a matrix is equal to its trace. So the sum of the eigenvalues of $A$ is $a+d$.

Answer 2

The eigenvalues satisfy the characteristic equation: $(\lambda-a)(\lambda-d)-bc=0$. Then $\lambda^2-(a+d)\lambda+ad-bc=0$.

But, $\lambda_2+a+b=a+d\implies \lambda_2=d-b$.

Answer 3

$$AU=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} =\begin{bmatrix} a+b \\ c+d \end{bmatrix}=(a+b) U,~ a+b=c+d.$$ so $\lambda_1=a+b$, if other eigenvalue is $\lambda_2$, then $\lambda_2=(a+d)-(a+b)=d-b$ (by the trace theorem). These two eigenvalues are also consid\stent with $\lambda_1 \lambda_2=\det A =ad-bc.$