$$U = \pmatrix{ a_1 & a_2\\ a_3 & a_4 } $$ $$ V = \pmatrix{ b_1 & b_2\\ b_3 & b_4 } $$

$$ \langle U,V\rangle = a_1b_1+a_2b_2+a_3b_3+a_4b_4 $$ $W= \{t(2, 0, 0, -1): t \in \Bbb R \text{ in } M_{2\times 2}\}$

Find the Basis and dimension of orthogonal complement of $W$ (that is, $W^\perp$)

I am having trouble finding the answer to this... Could someone please help me out?

  • $\begingroup$ I'll fix the formatting for this question. For future reference, see this tutorial on how to format mathematical writing. $\endgroup$ Oct 29, 2014 at 16:33
  • $\begingroup$ All right, 2 things I don't understand here: first, should $A_i,B_i$ and $T$ be lowercase? That is, are $a_1$ and $A_1$ the same thing? Second, what is M22 supposed to mean here? $\endgroup$ Oct 29, 2014 at 16:37
  • $\begingroup$ Also, since you are new to this website: it makes it significantly easier to answer questions when askers share what they've done so far, and questions are generally shut down when it appears that the asker has made no effort of his/her own. So, what have you tried so far? Do you understand what all of the words in the question mean? Do you know what $W^\perp$ should look like, or at least what dimension it should have? $\endgroup$ Oct 29, 2014 at 16:40
  • $\begingroup$ You can add a comment by clicking the "add a comment" button that appears below my comments here. $\endgroup$ Oct 29, 2014 at 16:42
  • $\begingroup$ Im sorry, Yes i am new to this website...M sub 2 2 for this question W perp is a sub space of V, intersects with W is only zero vector and (w perp)perp is W. $\endgroup$
    – Jane
    Oct 29, 2014 at 16:57

1 Answer 1


So, $W$ is the space given by $$ W = \left\{\pmatrix{2t&0\\0&-t}: t \in \Bbb R\right\} $$ What we are looking for is $W^\perp$, which by definition is $$ W^\perp = \left\{V \in M_{2 \times 2}: \langle U,V \rangle = 0 \text{ for every } U \in W \right\} $$ However, applying all of the above definitions, we can rewrite this statement as $$ W^\perp = \left\{\pmatrix{a_1 & a_2\\a_3 & a_4}: (2t)a_1 + (0)a_2 + (0)a_3 + (-t)a_4 = 0 \text{ for every } t \in \Bbb R \right\} $$ That is, we want to find the set of quadruples ("vectors") $(a_1,a_2,a_3,a_4)$ such that $$ 2ta_1 + -ta_4 = 0 \text{ for every } t \in \Bbb R $$ which is to say that we want the set of vectors $(a_1,a_2,a_3,a_4)$ such that $$ 2a_1 - a_4 = 0 $$ Equivalently, we want to find the solution set of the matrix equation $$ \pmatrix{2 & 0 & 0 & -1} \pmatrix{a_1\\a_2\\a_3\\a_4} = 0 $$ You should be able to find a basis consisting of $3$ elements.


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