# Finding the Eigenspace of a linear transformation

Let $$T:\mathbb{R}^2\to \mathbb{R}^2$$ be a linear transformation such that $$(a,b)\longmapsto (a+b, a-b)$$ Find all the eigenvalues and, for each eigenvalue, find the corresponding eigenspace.

My attempt:

I don't know if there is a normal procedure to find the matrix of a linear transformation, but I just "back filled" the entry values to make it work. So I have $$\begin{pmatrix} 1 & 1 \\ 1 & -1 \\ \end{pmatrix} \begin{pmatrix} a \\ b \\ \end{pmatrix}= \begin{pmatrix} a+b \\ a-b \\ \end{pmatrix}$$ So, denoting the matrix as $$A$$, I used the characteristic polynomial $$det(A-\lambda I)= \begin{pmatrix} 1-\lambda & 1 \\ 1 & -1-\lambda \\ \end{pmatrix}=0$$ $$\implies -(1-\lambda)^2-1=0\implies \lambda= 1+i$$ or $$1-i$$.

pluging the former value into the matrix I solve for $$\begin{pmatrix} -i & 1\\ 1 & -2-i \\ \end{pmatrix} \begin{pmatrix} a \\ b \\ \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ \end{pmatrix}$$ Which generates the system of equations $$-ia+b=0, \quad a-2b-ib=0$$ But solving the system gives me $$a=b=0$$.

There is a previous problem where I got the same thing, so I'm wondering if I am doing something wrong. Any help is much appreciated.

• $-1-\lambda \neq 1-\lambda$ – Chinnapparaj R Mar 6 at 8:01
• Note that the matrix is symmetric; the eigenvalues should be real! – Theo Bendit Mar 6 at 8:04

The eigen values are $$\pm \sqrt 2$$.
While computing $$det(A- \lambda I)$$. You are committing a mistake.