Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Q4. Suppose that $P$ is an $n \times n$ matrix such that $P^{2} = P$. Show that $\mathbb{R}^{n}$ is the direct sum of the range $R(P)$ and the nullspace $N(P)$ of $P$. Show also that $P$ represents the projection from $\mathbb{R}^{n}$ onto $R(P)$.

A4. Again – not sure how to start. Its idemptotent... so???

Q5. Prove that for any real matrix $A$, $N(A) = \text{orthogonal complement of } \ R(A^{T})$. Prove that for any real matrix, $N(A^{T}) = \text{orthogonal complement of} \ R(A)$.

A5. I get the first bit of this. It’s the second part I’m not sure about. Let $A^{\ast} = A^{T}$. Let $R(A)^{\ast}$ be the orthogonal complement of $R(A)$. What I said was let $x \in N(A)$. Then $Ax = 0$. So $A^{\ast}(Ax)=A^{\ast}0=0$. So $A^{\ast}(Ax)=0$. So $A(A^{\ast}x)=0$ so $A^{\ast}x=0$. So x is in the nullspace of $A^{\ast}$. I’m not sure about show to show the row space bit.

Q6. Let $A$ be a real matrix and let $R(A)$ denote its range. Show that the projection of a vector $b$ onto $R(A)$ parallel to the orthogonal complement of $R(A) – R(A)^{\ast}$ - is the vector in $R(A)$ closest to b.

Let A be a real matrix and let R(A) and R(A)* denote the range of A and the orthogonal complements of R(A) respectively. Show that the projection of a vector s onto R(A) parallel to R(A)* is the vector in R(A) closest to s.

A6. I think both of these questions are the same right? No idea where to start!!! :(

Im guessing these are all similar proofs - hence its on one thread.

share|cite|improve this question
Please ask one question per question, and preferrably not all at the same time. – Mariano Suárez-Alvarez May 20 '11 at 13:50
Also: there is rarely need of repeating question or exclamation marks... – Mariano Suárez-Alvarez May 20 '11 at 13:51
up vote 3 down vote accepted

Hint: For the first question, Observe that if $v\in R(P)$ then $Pv=v$ (why?). Deduce from here that $N(P)\cap R(P)=\{{0\}}$. Now use the theorem of dimensions for $N(P),R(P)$ and for sums and intersections.
Added after OP's comment: Did you prove the hint? If yes then take any $v\in N(P)\cap R(P)$. Then you have $0=P(v)=v$ from the hint. Hence $N(P)\cap R(P)=\{{0\}}$. Since $\dim N(P)+\dim R(P)=\dim \mathbb{R}^n=n$ and $$\dim(N(P) + R(P))=\dim N(P)+\dim R(P) - \dim(N(P) \cap R(P))=n$$ we have $N(P) + R(P)=\mathbb{R}^n$

share|cite|improve this answer
Thanks for the hint.... but I seem to be missing a step or two. Im still not getting the answer. I get when P(v)= - its a projection right? – user4645 May 23 '11 at 10:22
@user4645: I edited my answer – Dennis Gulko May 23 '11 at 17:22
hmmm.... any ideas on the other ones? :) – user4645 May 25 '11 at 10:37

Q5: Take any $v\in N(A)$ and $u\in R(A^T)$. We have $$\langle v,u \rangle = \langle v,A^T w \rangle= \langle Av,w \rangle = \langle 0,w \rangle=0$$ Which implies $N(A) \subseteq R(A^T)^\perp$.
For the other direction, take $v\in R(A^T)^\perp$. For any $u\in R(A^T)$ we have $0=\langle v,u \rangle$. Since for any $w \in V$ we have $A^Tw\in R(A^T)$, it follows that for any $w \in V$: $0=\langle v,A^Tw \rangle=\langle Tv,w \rangle$. Hence $Av\in V^\perp=\{{0\}}$ for all $w \in V$. Hence $v\in N(A)$. Hence $R(A^T)^\perp \subseteq N(A)$, which concludes the proof.
The other part follows easily by noticing that $(A^T)^T=A$ for finite-sized matrices.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.