[SOLVED] Prove that the only vector common is the zero vector

**Runty** · February 2nd 2010, 05:35 AM

Let $\lambda$ and $\mu$ be distinct eigenvalues for $T:R^n \rightarrow R^n$ and let the corresponding eigenspaces be $V_\lambda$ and $V_\mu$ respectively.

Prove that the only vector common to both $V_\lambda$ and $V_\mu$ is the zero vector.

**Opalg** · February 2nd 2010, 11:05 AM

Originally Posted by Runty

Let $\lambda$ and $\mu$ be distinct eigenvalues for $T:R^n \rightarrow R^n$ and let the corresponding eigenspaces be $V_\lambda$ and $V_\mu$ respectively.

Prove that the only vector common to both $V_\lambda$ and $V_\mu$ is the zero vector.

If the vector x is in both $V_\lambda$ and $V_\mu$ then $Tx = \lambda x$ and $Tx = \mu x$ . What can you deduce from that?

**Runty** · February 2nd 2010, 11:12 AM

Originally Posted by Opalg

If the vector x is in both $V_\lambda$ and $V_\mu$ then $Tx = \lambda x$ and $Tx = \mu x$ . What can you deduce from that?

That certainly helps a lot, but I suspect something is missing. Am I just to show that the two do not equal unless the zero vector is used, or is there something else I'd need to show?

**Opalg** · February 2nd 2010, 11:20 AM

Originally Posted by Runty

That certainly helps a lot, but I suspect something is missing. Am I just to show that the two do not equal unless the zero vector is used, or is there something else I'd need to show?

If $\lambda x = \mu x$ then $(\lambda-\mu)x = 0$ , and if $\lambda\ne\mu$ then $\lambda-\mu$ is a nonzero real number, so you can divide by it. That's all that is needed.

**Runty** · February 2nd 2010, 11:22 AM

Originally Posted by Opalg

If $\lambda x = \mu x$ then $(\lambda-\mu)x = 0$ , and if $\lambda\ne\mu$ then $\lambda-\mu$ is a nonzero real number, so you can divide by it. That's all that is needed.

Thanks, I had a feeling that's what I needed to do.

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