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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 $(A-\lambda_1 I)(A-\lambda_2 I)\cdots (A-\lambda_k I)x=0.$

Please help me to prove this.

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  • $\begingroup$ You could try by induction over $k$, starting with $k=3$. $\endgroup$ – Surb Jan 15 at 15:34
  • $\begingroup$ How the induction will work here as to how do we go for $k=2$ $\endgroup$ – MANI Jan 15 at 15:39
  • $\begingroup$ 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- $\endgroup$ – Surb Jan 15 at 15:53
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    $\begingroup$ 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. $\endgroup$ – JMoravitz Jan 15 at 16:01
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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.

$$\left[\prod_{j\ne 3}(A-\lambda_j I)\right](A-\lambda_3 I) x = 0.$$

Okay, after posting this, looked at the comments. @JMoravitz already gave the answer there!

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