$\begingroup$

What is the dimension of the vector space of all diagonal $n \times n$ matrices?

I tried to count how many distinct entries it has.

$\endgroup$ 5

1 Answer

$\begingroup$

Consider the set $\beta = \{D_1, \dotsc, D_n\}$ where $D_i$ is a matrix of all zero entries except $1$ in the $i^\text{th}$ entry along the diagonal. Then this is a basis of the vector space $V$ of all diagonal $n \times n$ matrices. To prove this we need to show $\operatorname{span}(\beta) = V$ and $\beta$ is linearly independent. The first is clearly true since $\operatorname{diag}(a_1, \dotsc, a_n) = a_1D_1 + \dotsb + a_nD_n$. Showing the second criteria is also easy. Suppose $$ c_1D_1 + \dotsb + c_nD_n = O $$ for some scalars $c_1, \dotsc, c_n$. Then this gives $\operatorname{diag}(c_1, \dotsc, c_n) = O$ and by definition of equality of two matrices, we must have $c_1 = \dotsb = c_n = 0$. So $\beta$ is a basis. Since it has $n$ elements, $\dim(V) = n$.

I hope this helps you understand the procedure in finding the dimension.

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy