Answer by user for Linear algebra change of basis always makes a linear map...
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...
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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...
View ArticleLinear algebra change of basis always makes a linear map diagonal.
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...
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