# How to show that $N(A) = R{(A^*)}^\perp$ and $N(A^*)=R({A})^\perp$?

How to show that for a given square matrices $N(A) = R{(A^*)}^\perp$ and $N(A^*)=R{(A)}^\perp$ where $N(A)$ and $R(A)$ are the null and range spaces of matrix $A$, respectively?

I am not able to figure out how to start?I find difficulty when I have to deal with the orthogonal complement of subspaces.

Thanks for helping me.

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The second statement must be $N(A^*) = R(A)^{\perp}$. – user17762 May 17 '12 at 5:38
@Marvis Thanks for pointing please edit that. – srijan May 17 '12 at 5:42

## 1 Answer

HINT Let $z \in N(A)$ and $x \in R(A^*)$. This gives us $Az = 0$ and $x = A^*b$, for some $b$. $x^*z = b^*Az = 0$. Hence, $x \perp z$.

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I need hint for second part too. Can you please help me out? – srijan May 17 '12 at 6:05
@srijan Replace $A$ by $A^*$ and make use of the fact that $(A^*)^* = A$ – user17762 May 17 '12 at 6:05
Let $z\in N(A)^*$ and $x\in R(A)$. This gives us $A^*z = 0$ and $x = Ab$, for some $b$. Further $x^*z = b^*A^*z = 0$. Hence, $x \perp z$. Am I correct? – srijan May 17 '12 at 6:10
@srijan yes.... – user17762 May 17 '12 at 6:13
Thanks a lot for helping me. – srijan May 17 '12 at 6:14