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.

  • 1
    $\begingroup$ Please ask one question per question, and preferrably not all at the same time. $\endgroup$ – Mariano Suárez-Álvarez May 20 '11 at 13:50
  • 1
    $\begingroup$ Also: there is rarely need of repeating question or exclamation marks... $\endgroup$ – Mariano Suárez-Álvarez May 20 '11 at 13:51

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 $N(P) + R(P)=\mathbb{R}^n$

  • $\begingroup$ 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? $\endgroup$ – user4645 May 23 '11 at 10:22
  • $\begingroup$ @user4645: I edited my answer $\endgroup$ – Dennis Gulko May 23 '11 at 17:22
  • $\begingroup$ hmmm.... any ideas on the other ones? :) $\endgroup$ – 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.


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