# Thread: Find a spanning set for the space

1. ## Find a spanning set for the space

Im not quite sure how to begin this problem

Find a spanning set for the space:

T(A) = {X in R^5 : AX = (A^T)X}

A =

29, 9, 11, -24, -6
9 , 46, -30, -30, -25
11, -30, 95, 36, -10
14, 13, -15, 57, -11
-6, -25, -10, 1, 28

Im also New to the forum and dont know how to properly post a matrix, sorry, But aside from that

How do i approach this problem? as far as i know to find the spanning set i find my reduced row-echelon form matrix, and Sort out the dependent and independent columns, but this matrix is rather large and ugly which makes me feel like theres a trick to this that involves MUCH less work than RREF.

2. Originally Posted by Sharpe116
Im not quite sure how to begin this problem

Find a spanning set for the space:

T(A) = {X in R^5 : AX = (A^T)X}

A =

29, 9, 11, -24, -6
9 , 46, -30, -30, -25
11, -30, 95, 36, -10
14, 13, -15, 57, -11
-6, -25, -10, 1, 28

Im also New to the forum and dont know how to properly post a matrix, sorry, But aside from that

How do i approach this problem? as far as i know to find the spanning set i find my reduced row-echelon form matrix, and Sort out the dependent and independent columns, but this matrix is rather large and ugly which makes me feel like theres a trick to this that involves MUCH less work than RREF.

More than ugly I'd say that matrix is horrible, nasty and mosntruous...and unfortunately for you you will have to take a general

vector X and check what the conditions on its entries are so that $\displaystyle AX = A^tX$ .

I can't see how finding the REF of A will help you at all...

Tonio

3. Could you elaborate on that a little? do you mean to make a system and solve for my X? To be honest im not sure How to use the condition AX = (A^T)X

4. $\displaystyle A=\left[\begin{array}{ccccc} 29 & 9 & 11 & -24 & -6\\ 9 & 46 & -30 & -30 & -25\\ 11 & -30 & 95 & 36 & -10\\ 14 & 13 & -15 & 57 & -11\\ -6 & -25 & -10 & 1 & 28 \end{array}\right]$

$\displaystyle A\vec{x}=A^{T}\vec{x}\rightarrow (A-A^T)\vec{x}=0$

$\displaystyle \vec{x}=\ker(A-A^T)=span( \left[\begin{array}{c} - \frac{43}{38}\\ 1\\ 0\\ 0\\ 0 \end{array}\right], \left[\begin{array}{c} \frac{51}{38}\\ 0\\ 1\\ 0\\ 0 \end{array}\right], \left[\begin{array}{c} \frac{6}{19}\\ 0\\ 0\\ 0\\ 1 \end{array}\right] )$

5. THAT is the answer, can you please explain how you calculated that answer? (A-A^T) i calculated to be:

$\displaystyle \left[\begin{array}{ccccc} 0 & 0 & 0 & -38 & 0\\ 0 & 0 & 0 & -43 & 0\\ 0 & 0 & 0 & 51 & 0\\ 38 & 43 & 51 & 0 & -12\\ 0 & 0 & 0 & 12 & 0 \end{array}\right]$

where do i go from here? ker(A-A^T) means finding the nullspace?

6. Yes, kernel is nullspace

rref : Reduced Row Echelon Form by Gauss-Jordan Elimination Operation

$\displaystyle \ker(M)=\ker(rref(M))$

$\displaystyle \ker\left[\begin{array}{ccccc} 0 & 0 & 0 & -38 & 0\\ 0 & 0 & 0 & -43 & 0\\ 0 & 0 & 0 & 51 & 0\\ 38 & 43 & -51 & 0 & -12\\ 0 & 0 & 0 & 12 & 0 \end{array}\right] \begin{array}{c} \div (-38) \\ -k_1\cdot R_1 \\ -k_2\cdot R_1 \\ \\ -k_3\cdot R_1 \end{array}=\ker\left[\begin{array}{ccccc} 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 38 & 43 & -51 & 0 & -12\\ 0 & 0 & 0 & 0 & 0 \end{array}\right]$

Now we solve $\displaystyle \ker\left[\begin{array}{ccccc} 0 & 0 & 0 & 1 & 0\\ 38 & 43 & -51 & 0 & -12 \end{array}\right]\rightarrow \left[\begin{array}{ccccc} 0 & 0 & 0 & 1 & 0\\ 38 & 43 & -51 & 0 & -12 \end{array}\right] \left[\begin{array}{c} x_1\\x_2\\x_3\\x_4\\x_5 \end{array}\right]=0$

$\displaystyle \rightarrow\left\{\begin{array}{rrrrrr} & & & x_4 & &=0 \\ 38x_1 & +43x_2 & -51x_3 & & -12 x_5&=0 \end{array}\right.$

$\displaystyle \rightarrow\left\{\begin{array}{ll} x_4 =0 \\ x_1 = \frac{1}{38}(-43x_2 +51x_3 +12 x_5) \end{array}\right.$ . So 2 dimensions are removed!

$\displaystyle \vec{x}=\left[\begin{array}{c} x_1\\x_2\\x_3\\x_4\\x_5 \end{array}\right]=\left[\begin{array}{c} \frac{-43x_2 +51x_3 +12 x_5}{38}\\x_2\\x_3\\0\\x_5 \end{array}\right]=x_2\left[\begin{array}{c} -\frac{43}{38}\\1\\0\\0\\0 \end{array}\right]+x_3\left[\begin{array}{c} \frac{51}{38}\\0\\1\\0\\0 \end{array}\right]+x_5\left[\begin{array}{c} \frac{12}{38}\\0\\0\\0\\1 \end{array}\right]$

$\displaystyle =span( \left[\begin{array}{c} - \frac{43}{38}\\ 1\\ 0\\ 0\\ 0 \end{array}\right], \left[\begin{array}{c} \frac{51}{38}\\ 0\\ 1\\ 0\\ 0 \end{array}\right], \left[\begin{array}{c} \frac{6}{19}\\ 0\\ 0\\ 0\\ 1 \end{array}\right] )$