# Find an ordered basis of $V$ such that $[T]_\beta$ is a diagonal matrix.

The entire problem statement is:

Let $V$ be a finite dimensional vector space and $T:V\to V$ be the projection of $W$ along $W'$, where $W$ and $W'$ are subspaces of $V$. Find an ordered basis $\beta$ of $V$ such that $[T]_\beta$ is a diagonal matrix.

The book I took this problem from is Linear Algebra by Friedberg, Insel, & Spence.

My attempt of the proof is as follows:

We have that $V=W\oplus W'$, so let $\beta_1=\{v_1,\dots,v_k\}$ and $\beta_2=\{v_{k+1},\dots,v_n\}$ be bases for $W$ and $W'$, respectively. Then we have that $\beta_1\cap\beta_2=\varnothing$ and $\beta=\beta_1\cup\beta_2$ is a basis for $V$. Now to determine the matrix representation of $T$ in the ordered basis $\beta$ consider $T(v_j)$ for $1\leq j\leq k$, \begin{align} T(v_j)&=\sum_{i=1}^ka_{ij}v_i\nonumber\\ &=1\cdot v_j\nonumber \end{align} The last equality occurs since $\beta_1$ is a basis for $W$ and hence linearly independent. Considering $[T]_\beta$ as a block matrix, it follows from above that the sub-matrix composed of the first $k$ rows and $k$ columns is the identity matrix of size $k$, that is, $I_k$. Moreover, since $T$ is the projection of $W$ along $W'$, the entries of $[T]_\beta$ can be entirely specified using only $\beta_1$. Hence, the entries outside of the $k\times k$ sub-matrix are zero. Thus we have that $$[T]_\beta=\left( \begin{array}{c|c} I_k & 0 \\ \hline 0& 0\\ \end{array}\right)$$ where the off diagonal zero matrices have size $(n-k)\times k$ and $k\times(n-k)$. Also, the diagonal zero matrix has size $(n-k)\times (n-k).$

My main concern is when I stated that $V=W\oplus W'$, which then allowed me to conclude that $\beta=\beta_1\cup\beta_2$ was a basis for $V$. The reason I'm assuming this is because the definition provided for the projection $T$ on $W$ along $W'$, it is assumed that $V$ is the direct sum of $W$ and $W'$.

Thanks for any help or feedback!

Well, you are correct that $V=W\oplus W'$, but a much simpler approach is to use the fact that for any projection we have $T^2=T$ so that $T(T-I)=0$. So in general we then have that the minimal polynomial of $T$ is $m_T(x)=x(x-1)$ (the special "trivial" cases being $T= 0$ so that $W=0$ and $T = I$ so that $W'=0$). Since this polynomial consist only of linear factors, $T$ is diagonalizable with eigenvalues $1$ and $0$.

Now we have $T(w)=w$ if and only if $w \in W$, so that the eigenspace associate with $1$ is $W$. Similarly we have $T(w')=0$ If and only if $w' \in W'$ so that the eigenspace associated with $0$ is $W'$. So any basis $\beta=\beta_W \cup \beta_{W'}$ where $\beta_W$ and $\beta_{W'}$ are bases for $W$ and $W'$ respectively is such that $[T]_\beta$ is diagonal.

• Thanks for the response. So far in the book, they have not discussed eigenvalues or eigenspaces. So far I only have the definition of a linear function, some associated theorems with regards to the null space and range. The section the problem is taken from introduces the idea of an ordered basis.
– S.D.
Apr 27, 2014 at 8:57
• ah ok I see... your approach looks sound, I would just explicitly show that $T(v_j)=0$ when $j>k$. And then you must make sure you know why $V=W\oplus W'$, maybe in answering the question you may assume it, but for yourself it would be good if you can work it out. You must prove that $V=W+W'$ and $W \cap W' = \{0\}$ OR that dim$(W)+$ dim$(W')=$ dim$(V)$. This last equation you can prove using the properties of the projection. Apr 27, 2014 at 9:31
• @ShantDanielian Are you using Friedberg as a textbook for a course? If so, is it postgrad? Or are you just self-studying? Apr 27, 2014 at 16:37
• Yes, I'm using Friedberg's textbook and yes I'm in a course using this book as well. Now, since $T$ is defined as the projection on $W$ along $W'$, and how I defined each basis, doesn't the fact $T(v_j)=0$ for $j>k$ just drop out? Moreover, I went through a problem which stated that if the intersection of two bases of a subspace are empty and their union make a basis for the parent vector space, then $V=W\oplus W'$. Moreover, from the definition of projection $T$, it assumes the fact V is written as a direct sum. Isn't that the case here?
– S.D.
Apr 27, 2014 at 20:55
• @ShantDanielian ok a few subtle problems here...I will refer to your original question: `Moreover, since $T$ is the projection of $W$ along $W′$, the entries of $[T]_\beta$ can be entirely specified using only $\beta_1$.' This statement is quite simply false. A zero entry isn't nothing. You have shown that the first $k$ columns of $[T]_\beta$ are the columns of $I_k$, but you have to show what the last $n-k$ columns are too. Regarding the moreover part: where did you prove that $\beta_1 \cap \beta_2 = \emptyset$? And sure it implies that $V$ is written as a direct sum, but do you know why? Apr 28, 2014 at 5:43