# Product of terms involving eigenvalues and an eigenvector.

Let $$A_{n\times n}$$ be a matrix and $$\lambda_1,\lambda_2\ldots \lambda_k$$ be $$k$$ eigenvalues of $$A$$. Let $$x$$ be an eigenvector of $$A$$ corresponding to eigenvalue $$\lambda_3$$. Then without expanding prove that $$(A-\lambda_1 I)(A-\lambda_2 I)\cdots (A-\lambda_k I)x=0.$$

• You could try by induction over $k$, starting with $k=3$. – Surb Jan 15 at 15:34
• How the induction will work here as to how do we go for $k=2$ – MANI Jan 15 at 15:39
• If $k=2$, your problem makes no sense since $\lambda_3$ does not exist. So you prove it for $k=3$, and then show for $k\geq 3$ with an induction step- – Surb Jan 15 at 15:53
• Does $(A-\alpha I)$ commute with $(A-\beta I)$? That is, does $(A-\alpha I)(A-\beta I) = (A-\beta I)(A-\alpha I)$? Use this result then to show that $(A-\lambda_1 I)(A-\lambda_2 I)(A-\lambda_3 I)\cdots (A-\lambda_k I)x$ $= (A-\lambda_1 I)(A-\lambda_2 I)(A-\lambda_4 I)\cdots (A-\lambda_k I)(A-\lambda_3 I)x$, having moved the special term in the product we care about to the end so it can act on $x$ first. – JMoravitz Jan 15 at 16:01

Assume $$n\ge 3$$ so that eigenvalue $$\lambda_3$$ exists.
$$(A-\lambda_3 I)x = 0$$ by definition of eigenvalue and eigenvector.
$$A$$ and $$I$$ commute, so we can rearrange the terms.
$$\left[\prod_{j\ne 3}(A-\lambda_j I)\right](A-\lambda_3 I) x = 0.$$