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Prove that there exist bases $\alpha $ and $\beta$ for V such that $ [T]_{\alpha}^{\beta} $ is a diagonal matrix with each diagonal entry equal to either 0 or 1.

Originally i thought that T=I was the only solution to this i realize that is not the case now but i am still lost what if T forms an non-invertable matrix? i really dont understand how we can always know that this is true cause if the map is non-invertable isn't it not diagonalizable?

Faust
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    Non-invertible matrices can also be diagonalizable. Consider the zero matrix. It is diagonal but singular. In fact, a diagonalizable matrix is singular iff at least one of the diagonal entries is zero (so the determinant is zero). – Levent Feb 07 '18 at 20:52
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    my concern wasn't does there exist a diagonalizable matrix that is non-invertable more of surprise that this question seem to imply every matrix formed from a linear map IS diagonalizable invertability notwithstanding – Faust Feb 07 '18 at 20:54
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    Often when people talk about diagonalization they refer to diagonalization with respect to a single basis. This is not the case here, where you are allowed to have two bases. – asdq Feb 07 '18 at 20:56
  • What is $[T]^b_a$? – Serge Seredenko Feb 07 '18 at 22:46
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    The groups of invertible matrices $\mathsf{GL}(n)$ and $\mathsf{GL}(m)$ act to sides on $\mathsf{Mat}(m,n)$, and you are considering two matrices $A$ and $B$ to be equivalent if for some $P$ and $Q$ in the corresponding groups, $A = PBQ$. When people talk about diagonalization, $m=n$, and two matrices are equivalent if $A = PBP^{-1}$ or some $P\in \mathsf{GL}(n)$. In the first case you are counting double cosets, and in this case the only invariant is the rank. Doing this takes about one or two classes. In the second you're looking at diagonalization and Jordan forms, which may take a quarter. – Pedro Feb 08 '18 at 00:41

2 Answers2

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Suppose that $T$ has rank k, then you can construct a basis made of $n-k$ vectors in Ker(T) and of $k$ vectors orthogonal to them

$$\alpha=\{\alpha_1,...\alpha_k,\alpha_{k+1},...\alpha_n\}$$

then assume

$$\beta=\{\beta_1,...\beta_k,\beta_{k+1},...\beta_n\}$$

with $\beta_i=T(\alpha_i)$ thus $[T]_{\alpha}^{\beta}$ is a diagonal matrix with each diagonal entry equal to either 0 or 1

user
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Let $T:V\rightarrow V$ be a linear map and $\alpha=\left\{v_1, \dots, v_k, u_{k+1}, \dots u_n\right\}$ a basis of $V$ such that $\left\{v_1, \dots, v_k\right\}$ is a basis of $\ker(T)$.

You can easily prove that $T(u_{k+1}), \dots , T(u_n)$ are linearly independent vectors. (Mimick the technique used to prove the rank-nullity theorem).

Define $\beta=\left\{w_1, \dots, w_k,T(u_{k+1}), \dots , T(u_n) \right\}$. Here $w_1, \dots, w_k$ are vectors extending the linearly independent $T(u_{k+1}), \dots , T(u_n)$ to a basis of $V$. Now you can easily check that $[T]_{\alpha}^{\beta}$ is of the required form.

Notice that this does not imply that $\det(T)=0$ or $1$. Indeed, $\det(T)$ is defined as $\det([T]_{\gamma}^{\gamma})$ for some basis $\gamma$ of $V$. So the matrix of $T$ is expressed w.r.t. the same basis in both domain and target.

Mathematician 42
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  • Excellent answer thank you! – Faust Feb 07 '18 at 21:04
  • Sure, no problem. Many students which are good at linear algebra get confused by this question. Indeed, it would be very weird that any linear map can be represented by a diagonal matrix if you spend so much time on diagonalization of matrices. But obviously diagonalization only makes sense w.r.t. the same basis. Otherwise eigenvectors would be a very common thing :) – Mathematician 42 Feb 07 '18 at 21:10