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I am self-studying linear algebra from the book Linear Algebra done right 2nd ed. These are 3 statements about UTM:

Suppose (an operator) T $\in$ L(V, V) and $(v_1, v_2, \ldots, v_n)$ is a basis of V. The following statements are equivalent:

  1. the matrix M(T) with respect to the basis $(v_1, \ldots, v_n)$ is upper triangular;
  2. $Tv_k \in span(v_1, \ldots, v_k)$ for each $k=1, 2, \ldots, n$;
  3. $span(v_1, \ldots, v_k)$ is invariant under T for each $k=1, 2, \ldots, n$;

The book says "The equivalence of 1 and 2 follows easily from the definition since 2 implies that the matrix elements below the diagonal are zero". I really can't figure out why 2 implies 1 or vice versa. Suppose $Tv_k \in span(v_1, \ldots, v_k)$. Then $Tv_1 = a_1v_1$, $Tv_2 = b_1v_1 + b_2v_2$, $\ldots$. When applied the operator to polynomial, Fundamental Theorem of Algebra says every operator has at least an eigenvalue, so $Tv_1 = a_1v_1$ gives no info about the matrix. How do we get to the conclusion that T must be upper triangular from this point? Thanks.

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The matrix $M(T)$ has as its columns the representation (in your basis) of $T(v_1), T(v_2), \ldots$. That is, the first column is the representation of $T(v_1)$. By hypothesis, $T(v_1)=a_1v_1=a_1v_1+0v_2+0v_3+\cdots+0v_n$. Hence the first column of $M(T)$ has $a_1$ in the first entry, then zero in all subsequent entries. Similarly for the other columns -- there are possibly nonzero entries above the diagonal, but zeroes below the diagonal because $T(v_k)=a_1'v_1+\cdots+a_k'v_k+0v_{k+1}+\cdots+0v_n$.

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