Question

Let u in R(n) be a nonzero row vector.
(a) show that the n x n matrix A = u(t)u is symmetric and that rank(A)=1
(b)Show that the matrix P=I(n)+u(t)u is invertible.

Answer 1

if u is just a row vector, then how is u(t)u anything but a single entry and the rank(A)=1, or do I just not understand what is going on here.
@helder_edwin

Answer 2

if u is a row vector then
\[ \large u=(u_1\quad u_2\quad\dotsb\quad u_n) \]
so \(u^t\cdot u\) is the matrix
\[ \large A=\begin{pmatrix}
u_1\\ u_2\\ \vdots\\ u_n
\end{pmatrix}\cdot\begin{pmatrix}
u_1 & u_2 & \dotsb & u_n
\end{pmatrix} \]
is a \(n\times n\) matrix

Answer 3

symmetry is easy
\[ \large A^t=(u^tu)^t=u^t(u^t)^t=u^tu=A \]

Answer 4

i have to go.
sorry.
something urgent came up.

Answer 5

I guess I was thinking about it as the row vector being first, then the column vector., I got symmetry that way, so then how is the rank =1

Answer 6

okay, thank you.

Answer 7

sorry i had to go.

when u have A=BC then rank(A)<=rank(B) and rank(A)<=rank(C).

Answer 8

this means that
\[ \large \text{rank}(u^tu)\leq\text{rank}(u)=1 \]
because u is a row vector.

Answer 9

now since the only matrix with rank zero is the zero matrix then
\[ \large \text{rank}(u^tu)>0 \]
therefore rank(u^tu)=1.

Answer 10

where did you get the rank(A)<=rank (B)? is that something that is just known to be true. A B and C are matrices?

Answer 11

yes. it is a theorem of linear algebra.

Answer 12

and the rank of u, would always 1 right, and since the zero matrix is the only rank 0, then rank(u(t)u) has to be 1.

Answer 13

yes

Answer 14

because \(u\neq0\)

Answer 15

okay, so now for part (b), when I add the identity to A, I still get A right.. and since A is symmetric, then it has to be invertible too?

Answer 16

no. u add 1 to every entry in the diagonal of A.

Answer 17

ohh yes, right, and because its just the diagonal the only thing that changes, it would be invertible.

Answer 18

i don't know how to prove that.

Answer 19

perhaps using induction on the dimension n of the matrix????

Answer 20

I thank you for your help and coming back to help with the problem. I will try to see if I can use induction.

Answer 21

there is a theorem that says
\[ \large \text{rank}(A+B)\leq\text{rank}(A)+\text{rank}(B) \]

Answer 22

so rank(P)<=n+1

Answer 23

if u use this u should have to disprove rank(P)=n+1 which is inmediat. since P is n*n.
so rank(P)<=n

Answer 24

now u would have to disprove rank(P)<n.

Answer 25

there is a theorem in this section that I did not know how to apply... maybe it helps, i don't really know.
its says.... nullity(A)+rank(A)=n

Answer 26

right. i forgot that. so another strategy would be to prove that nullity(P)=0

Answer 27

and why is the dim of the null space 0?

Answer 28

is has to be. if so, then rank(P)=n and P would be invertible.

Answer 29

okay, right b/c of that theorem, nullity(A) has to be zero, but if that is the case then why cant you just say its invertible b/c of the theorem?

Answer 30

i mean. u should prove that nullity(P)=0. if u get this, then P would be invertible because rank(P) would equal n.

Answer 31

i didn't express well my thoughts. sorry.

Answer 32

I guess i am stuck on how to prove the dim=0

Answer 33

if there is \(x\neq0\) such that Px=0 then
\[ \large 0=Px=(I+u^tu)x=Ix+(u^tu)x=x+(u^tu)x \]
so
\[ \large (u^tu)x=-x \]

Answer 34

It makes sense when I see it, I just can't figure out how to get things like this started. I appreciate your help though, I learn more for here than I do from the actual book.

Answer 35

i can't see why this last equation should be wrong.

Answer 36

sorry.
i'm very tired.

Answer 37

good luck with this problem.

Answer 38

its okay, you have been a great help. i thank you again for everything.