# Thread: Orthogonal vectors & column and null space

1. ## Orthogonal vectors & column and null space

I need help with the following questions.

1) show that X & Y are orthogonal in Rn if and only if ||x+y|| = ||x-y||
2)let A be an n x n matrix, find a matrix A for which col A = null A
3) if A is m x n matrix and B is n x m matrix, show that AB = 0 if and only if col B is contained in null A

1) if x & y are orthogonal then x * y = 0 then what?

3) let y be a vector in of B [y1...yn]T then BX = 0 for all X in Rn
col B {y | y= BX}

Y= BX
AY=ABX
but AB = 0

AY = 0X
then wat?

2. Originally Posted by amitmeh
1) show that X & Y are orthogonal in Rn if and only if ||x+y|| = ||x-y||
If $x=(x_1,...,x_n),y=(y_1,...,y_n)$. If $x\cdot y=0$ then $|x+y| = |x|+|y|$. Therefore, $|x+y| = |x|+|y|$ and $|x-y| = |x|+|-y|=|x|+|y|\implies |x+y| = |x-y|$. Conversely, if $|x+y| = |x-y|\implies |x+y|^2 = |x-y|^2$, which means $(x_1+y_1)^2 + ... + (x_n+y_n)^2 = (x_1-y_1)^2 + ... + (x_n-y_n)^2$. Open parenthesis and cancel, $2x_1y_1 + ... + 2x_ny_n = -2x_1y_1 - ... - 2x_ny_n$. Thus, $4(x_1y_1+...+x_ny_n) = 0 \implies -4(x\cdot y) = 0 \implies x\cdot y =0$

3. Originally Posted by amitmeh
2)let A be an n x n matrix, find a matrix A for which col A = null A
This can only be true if $n$ is even. Because by the rank-nullity theorem the rank of a matrix plus its nullity must be n. However, if the column space and nullspace are equal then it means the rank and nullity are the same. And so we have an even number for n.

Let $A$ be the following matrix: $a_{ij} = 0$ everywhere, except with $a_{ij}=1$ for $i=1,...,n/2$ and $j=n/2+i$.
For example, if $n=4$ then the matrix is:
$A=\begin{bmatrix}0&0&1&0\\0&0&0&1\\0&0&0&0\\0&0&0& 0\end{bmatrix}$

Notice that $A^2 = \bold{0}$. As a consequence we see that that the coloumn space is contained in the nullspace. To see this let $\bold{y}$ be in the coloumn space. Then $A\bold{x} = \bold{y}$ for some $\bold{x}\in \mathbb{R}^n$. Therefore, $A^2\bold{x} = A\bold{y} \implies A\bold{y} = \bold{0}$. Therefore, $\bold{y}$ is in the nullspace. To show that the coloumn space is equal to the null space it sufficies to show that they have the same dimension. The null space has dimension $n/2$, that is easy to see. Because the matrix is in row-reduced echelon form and so there are $n/2$ "free" variables that come up in solving the homogenous equation. Immediately we know that that coloumn space has dimension $n-n/2 = n/2$ and so this matrix is one such example that you are looking for.

3) if A is m x n matrix and B is n x m matrix, show that AB = 0 if and only if col B is contained in null A