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I'm trying to prove that symmetric matrix (with real entries) is diagonalizable. Here, Ittay Weiss proved the result. I'm following with his argument, but I couldn't properly understand the following part.

Comment: To triangulate the matrix, use induction of the order of the matrix. For $1\times 1$ it's trivial. For $n\times n$, first find any arbitrary eigenvector $v_1$ (one such must exist). Thinking of the matrix as a linear transformation on a vector space $V$ of dimension $n$, write $V$ as $V=V_1\oplus W$, where $V_1$ is the subspace spanned by $v_1$. Then $W$ is $n-1$-dimensional, apply the induction hypothesis to $A|_{W}$ to obtain a base $v_2,\ldots, v_n$ in which $A|_W$ is triangular. It now follows that in the base $v_1,\ldots, v_n$ $A$ is triangular.

I'm having trouble understanding the induction part. Especially, what does $A|_W$ mean? Any help would be thankful!

Kim
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$A$ is a linear transformation from $V$ to $V$, so it is in particular, a function from $V$ to $V$. $A\vert_W$ is simply the restriction of the function to the subspace $W$.

Kenneth Yeo
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  • No, in the reference $A$ is a matrix, and restricting a matrix to a subspace is really problematic (I left a comment at the referred answer, which is really utterly wrong). Even in case of a linear transformation, you need to have a stable subspace in order to restrict and get a linear transformation $W\to W$, and nothing in the given formulation there ensures that $W$ is stable. – Marc van Leeuwen Aug 21 '19 at 15:09