0
$\begingroup$

Find A a 3x3 real symmetric matrix such that the linear map A: R3 to R3 with x to Ax satisfies null(A) = {a(1,1,1) : a in R}

If anyone has a slow detailed explanation, I'd appreciate it!

https://i.stack.imgur.com/doSKG.png

$\endgroup$

closed as off-topic by Adrian Keister, RRL, Andrew, Shaun, saz Feb 5 at 21:17

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, RRL, Andrew, Shaun, saz
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Can you at least construct any matrix that has the correct null space? $\endgroup$ – amd Feb 5 at 1:45
  • $\begingroup$ Not really, how would I go about this? $\endgroup$ – blo Feb 5 at 2:13
1
$\begingroup$

The matrix $$B=\pmatrix{1&-1&0\\0&1&-1\\}$$ has null-space $$\Bbb R\pmatrix{1\\1\\1}.$$ The matrix $A=B^tB$ is symmetric, and has the same null-space.

$\endgroup$
  • $\begingroup$ how would i construct the matrix given the null? it seems like you just magically came up with one. $\endgroup$ – blo Feb 6 at 19:42
  • $\begingroup$ Basically, it is the solution space to a system of homogeneous linear equations, so you could use row reduction.... $\endgroup$ – Lord Shark the Unknown Feb 6 at 19:43
  • $\begingroup$ By attempt, was setting the null space vector = 0, so i got ax1 = ax2 = ax3 = 0. and I solved for each x vector. Can you tell me where am i approaching this incorrectly? $\endgroup$ – blo Feb 6 at 19:57

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