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Math Help - Upper-Triangular Matrix - Invertible Operator Theorem

  1. #1
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    Upper-Triangular Matrix - Invertible Operator Theorem

    I am trying to understand the proof of this theorem which states

    Suppose T\in\mathcal{L}(V) has an upper-triangular matrix with respect to some basis in V. Then T is invertible if and only if all the entries on the diagonal of that upper-triangular matrix are nonzero.

    I understand the proof in the first direction. However I am a bit lost on proving that the converse is true.

    The proof follows:

    Suppose that T is not invertible. Thus T is not injective, and hence there exists a nonzero vector v ∈ V such that Tv = 0. Because (v_1,\dots,v_n) is a basis of V, we can write v=a_1v_1+\cdots+a_kv_k

    where a_1,\dots,a_k\in\mathbb{F} and a_k\ne0 (represent v as a linear combination of (v_1,\dots,v_n) and then choose k to be the largest index with a nonzero coefficient). Thus

    0=Tv
    0=T(a_1v_1+\cdots+a_kv_k)
    0=T(a_1v_1+\cdots+a_{k-1}v_{k-1})+a_kTv_k

    I understand everything so far until they say

    The last term in the parentheses is in span(v_1,\dots,v_{k-1}). Thus the last equation shows that a_kTv_k\inspan(v_1,\dots,v_{k-1})

    I don't quite understand why this is true.

    Thank you.
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  2. #2
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    Quote Originally Posted by Anthonny View Post
    I am trying to understand the proof of this theorem which states

    Suppose T\in\mathcal{L}(V) has an upper-triangular matrix with respect to some basis in V. Then T is invertible if and only if all the entries on the diagonal of that upper-triangular matrix are nonzero.

    I understand the proof in the first direction. However I am a bit lost on proving that the converse is true.

    The proof follows:

    Suppose that T is not invertible. Thus T is not injective, and hence there exists a nonzero vector v ∈ V such that Tv = 0. Because (v_1,\dots,v_n) is a basis of V, we can write v=a_1v_1+\cdots+a_kv_k

    where a_1,\dots,a_k\in\mathbb{F} and a_k\ne0 (represent v as a linear combination of (v_1,\dots,v_n) and then choose k to be the largest index with a nonzero coefficient). Thus

    0=Tv
    0=T(a_1v_1+\cdots+a_kv_k)
    0=T(a_1v_1+\cdots+a_{k-1}v_{k-1})+a_kTv_k

    I understand everything so far until they say

    The last term in the parentheses is in span(v_1,\dots,v_{k-1}).

    Thus the last equation shows that a_kTv_k\inspan(v_1,\dots,v_{k-1})
    Do you mean that a_kTv_k is in the span of (v_1, \cdots, v_{k-1})?
    That follows from the fact that a_kTv_k= T(a_1v_1+ \cdots+ a_{k-1}v_{k-1}) and the right side is in the span.

    I don't quite understand why this is true.

    Thank you.
    A simpler proof is that a matrix is invertible if and only if its determinant is not 0. And the determinant of a triangular matrix is the product of the numbers on the diagonal.
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