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If we have two square matrices of the same order n over the field $F$ and the same rank, is there a linear transformation from $F^n$ to $F^n$ such that the two matrices are the representation of it in some basis (not necesary the same basis of domain and codomain) ?

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I don't understand the question. What "two matrices"? Representation of what? –  wj32 Nov 7 '12 at 8:32
    
the question is if you can define a linear transformation that the matrices a re matrix representation of it in some basis.If the matrices ar similar is a wellknown result. If not similsr? –  tota Nov 7 '12 at 8:38

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$\def\rank{\operatorname{rank}}\def\Mat{\operatorname{Mat}}\def\GL{\operatorname{GL}}\def\im{\operatorname{rng}}$Let $A \in \Mat_n(F)$ with $\rank A = k$. Let $a_1, \ldots, a_k$ be a basis of $\im A$, complete these vectors to a basis $a_1, \ldots, a_n$ of $F^n$ and let $S = (a_1\mid \cdots \mid a_n) \in\GL_n(F)$ (columnswise). Let $\alpha_{k+1}, \ldots, \alpha_{n}$ a basis of $\ker A$, choose $\alpha_i$ for $1 \le i \le k$ such that $A\alpha_i = a_i$. Then $\alpha_1, \ldots, \alpha_n$ is a basis of $F^n$ and set $T = (\alpha_1\mid\ldots\mid \alpha_n) \in \GL_n(F)$. Now define $C_k := S^{-1}AT$, for $1 \le i \le k$ we have \[ C_ke_i = S^{-1}ATe_i = S^{-1}A\alpha_i = S^{-1}a_i = e_i \] and for $k+1 \le i \le n$ \[ C_ke_i = S^{-1}ATe_i = S^{-1}A\alpha_i = 0 \] So \[ C_k = \begin{pmatrix} 1 & 0 & \cdots & 0\\ 0 & 1 & \cdots & 0 \\ 0 & 0 & \ddots & \vdots \\ 0 & 0 & & 0\end{pmatrix} \] with $k$ ones on the diagonal.

Now let $A, B \in \Mat_n(F)$ with $\rank A = \rank B = k$ be given. Find $S_1, T_1, S_2, T_2 \in \GL_n(F)$ with $S_1AT_1 = S_2BT_2 = C_k$. Hence in the bases given by the $T_i$ and $S_i$ both $A$ and $B$ represent a projection onto the first $k$ coordinates.

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NICE PROOF, THANK YOU! –  tota Nov 7 '12 at 12:04

The answer is no. For example if $F$ is a field with more than two elements, then the matrix $\begin{pmatrix} 1&0\\0&1\end{pmatrix}$ can represent only the trivial transformation, while $\begin{pmatrix} 2&0\\0&2\end{pmatrix}$ can represent only th transformation the double each vector.

In general, two matrices $A,B$ represent the same linear transformation in different bases if and only if $A$ is similar to $B$, i.e. there exists an invertible matrix $M$ such that $A=M^{-1} B M$. Indeed, $M$ will be the transition matrix from one basis to another.

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If I understand you are speaking about taking the same basis of the domain and codomain.Then you are wright. But my question is also taking any basis ,no mind if you take the same basis. –  tota Nov 7 '12 at 12:17
    
So I misunderstood your question. Sry –  Lior B-S Nov 7 '12 at 19:46

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