# Thread: [SOLVED] Prove that the only vector common is the zero vector

1. ## [SOLVED] Prove that the only vector common is the zero vector

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.

2. 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?

3. 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?

4. 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.

5. 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.