# Find the Basis and dimension of orthogonal complement of W

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

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

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

• 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? Commented Oct 29, 2014 at 16:37
• 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? Commented Oct 29, 2014 at 16:40
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.