# subtracting the eigenvalue matrix from the original transformation of a 2x2 hit's the other eigenvalue proof

Here is the question: Prove that if λ, μ are real eigenvalues of a 2 × 2 matrix, then any nonzero column of the matrix A − λI is an eigenvector for μ. I assume A is the linear map for which these are the eigenvalues. I am pretty lost after playing around with a generalized 2x2 for awhile

• What do you mean by "non-zero column"? – BigbearZzz Jan 15 '16 at 7:49
• your guess is as good as mine. – qbert Jan 15 '16 at 7:55

Write $A-\lambda I$ as an augmented matrix $[x,y]$. So, the statement "any nonzero column of the matrix $A − \lambda I$ is an eigenvector for $\mu$" means that $Ax=\mu x$ and $Ay=\mu y$ (regardless of whether $x$ or $y$ is zero). In turn, this means $A[x,y]=\mu[x,y]$, or $A(A-\lambda I)=\mu (A-\lambda I)$. In other words, what you are going to prove is that $$A^2-(\lambda+\mu)A + \lambda\mu I = 0,$$ or equivalently, $$A^2-\operatorname{trace}(A)A+\det(A)I=0.$$ You may verify this equality directly by explicitly calculating the LHS of $(1)$ in terms of the entries of $A$. This equality is actually a special case of Cayley–Hamilton theorem.

Start with a matrix $A = \left(\begin{array}{cc} a && b \\ c && d \end{array}\right)$, the eigenvalues are the roots of $(a - x)(d - x) - bc$.

We can write that better as $x^2 - (a + d)x + (ad - bc) = 0$, so we know $\mu + \lambda = a + d$ and $\mu\lambda = ad - bc$, in fact, this is just standard trace and determinant equalities.

Now suppose the first column of $A - \lambda I$ is non zero, so we have the vector $v = \left(\begin{array}{c}a - \lambda \\ c \end{array}\right)$

To verify it is an eigenvector, we multiply it with the matrix and get

$Av = \left(\begin{array}{c}a(a - \lambda) + bc \\ c(a-\lambda) + cd \end{array}\right) = \mu\left(\begin{array}{c}a - \lambda \\ c \end{array}\right)$

To justify the last equality, we write

$\begin{eqnarray*} a(a - \lambda) + bc &=& \mu(a - \lambda) \\ a^2 - a\lambda + bc &=& a\mu - \mu\lambda \\ a^2 + bc &=& a\mu + a\lambda - \mu\lambda \\ a^2 + bc &=& a(\mu + \lambda) - \mu\lambda \\ a^2 + bc &=& a(a+d) - (ad-bc) \\ a^2 + bc &=& a^2 + ad - (ad-bc) \\ 0 &=& 0 \end{eqnarray*}$

Of course, we read that backward.

All other cases can be proved similarly, lot of algebra.