Prove conditions for upper triangular matrix
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In Sheldon Axler's Linear Algebra Done Right, I find the proof of definition 5.26 to be insufficiently rigorous. It states the following:
$textbf{Conditions for upper-triangular matrix:}$
Suppose $T in mathscr L(V)$ and $v_1,...,v_n$ is a basis of $V$. Then the following are equivalent:
(a) the matrix of $T$ with respect to $v_1,...,v_n$ is upper-triangular
(b) $Tv_j in span(v_1,...,v_j)$ for each $j=1,...,n$
(c) $span(v_1,...,v_j)$ is invariant under $T$ for each $j=1,...,n$
In typical Sheldon Axler fashion, his proof begins as follows:
"The equivalence of (a) and (b) follows easily from the definitions and a moment's thought. Obviously (c) implies (b). Hence to complete the proof, we need only prove that (b) implies (c)......"
While I do see why this is intuitively true, can anyone provide more truly rigorous (and less condescending) proofs of (a)$Rightarrow$(b) and (c)$Rightarrow$(b)?
linear-algebra eigenvalues-eigenvectors
asked Dec 16 '18 at 7:53
jmarsjmars
22
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add a comment |
$begingroup$
In Sheldon Axler's Linear Algebra Done Right, I find the proof of definition 5.26 to be insufficiently rigorous. It states the following:
$textbf{Conditions for upper-triangular matrix:}$
Suppose $T in mathscr L(V)$ and $v_1,...,v_n$ is a basis of $V$. Then the following are equivalent:
(a) the matrix of $T$ with respect to $v_1,...,v_n$ is upper-triangular
(b) $Tv_j in span(v_1,...,v_j)$ for each $j=1,...,n$
(c) $span(v_1,...,v_j)$ is invariant under $T$ for each $j=1,...,n$
In typical Sheldon Axler fashion, his proof begins as follows:
"The equivalence of (a) and (b) follows easily from the definitions and a moment's thought. Obviously (c) implies (b). Hence to complete the proof, we need only prove that (b) implies (c)......"
While I do see why this is intuitively true, can anyone provide more truly rigorous (and less condescending) proofs of (a)$Rightarrow$(b) and (c)$Rightarrow$(b)?
linear-algebra eigenvalues-eigenvectors
asked Dec 16 '18 at 7:53
jmarsjmars
22
$endgroup$
add a comment |
$begingroup$
In Sheldon Axler's Linear Algebra Done Right, I find the proof of definition 5.26 to be insufficiently rigorous. It states the following:
$textbf{Conditions for upper-triangular matrix:}$
Suppose $T in mathscr L(V)$ and $v_1,...,v_n$ is a basis of $V$. Then the following are equivalent:
(a) the matrix of $T$ with respect to $v_1,...,v_n$ is upper-triangular
(b) $Tv_j in span(v_1,...,v_j)$ for each $j=1,...,n$
(c) $span(v_1,...,v_j)$ is invariant under $T$ for each $j=1,...,n$
In typical Sheldon Axler fashion, his proof begins as follows:
"The equivalence of (a) and (b) follows easily from the definitions and a moment's thought. Obviously (c) implies (b). Hence to complete the proof, we need only prove that (b) implies (c)......"
While I do see why this is intuitively true, can anyone provide more truly rigorous (and less condescending) proofs of (a)$Rightarrow$(b) and (c)$Rightarrow$(b)?
linear-algebra eigenvalues-eigenvectors
asked Dec 16 '18 at 7:53
jmarsjmars
22
$endgroup$
In Sheldon Axler's Linear Algebra Done Right, I find the proof of definition 5.26 to be insufficiently rigorous. It states the following:
$textbf{Conditions for upper-triangular matrix:}$
Suppose $T in mathscr L(V)$ and $v_1,...,v_n$ is a basis of $V$. Then the following are equivalent:
(a) the matrix of $T$ with respect to $v_1,...,v_n$ is upper-triangular
(b) $Tv_j in span(v_1,...,v_j)$ for each $j=1,...,n$
(c) $span(v_1,...,v_j)$ is invariant under $T$ for each $j=1,...,n$
In typical Sheldon Axler fashion, his proof begins as follows:
"The equivalence of (a) and (b) follows easily from the definitions and a moment's thought. Obviously (c) implies (b). Hence to complete the proof, we need only prove that (b) implies (c)......"
While I do see why this is intuitively true, can anyone provide more truly rigorous (and less condescending) proofs of (a)$Rightarrow$(b) and (c)$Rightarrow$(b)?
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
asked Dec 16 '18 at 7:53
jmarsjmars
22
asked Dec 16 '18 at 7:53
jmarsjmars
22
asked Dec 16 '18 at 7:53
jmarsjmars
22
asked Dec 16 '18 at 7:53
jmarsjmars
22
asked Dec 16 '18 at 7:53
jmarsjmars
22
22
add a comment |
add a comment |
1 Answer
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$a$ implies $b$ The $j$-column of the matrix of $T$ with respect to $v_1,...,v_n$ are the cordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$. Since the matrix of $T$ with respect to $(v_1,...,v_n)$ is upper-triangular, this implies that for $i>j$, the $i$-coordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$ is $0$, this implies that $T(v_j)in span(v_1,...,v_j)$.
$c$ implies $b$ if $Span(v_1,...,v_j)$ is invariant by $T$, it implies that $T(v_j)in span(v_1,...,v_j)$ since $v_jin (v_1,...,v_j)$
answered Dec 16 '18 at 8:06
Tsemo AristideTsemo Aristide
59.1k11445
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1 Answer
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1 Answer
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$begingroup$
$a$ implies $b$ The $j$-column of the matrix of $T$ with respect to $v_1,...,v_n$ are the cordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$. Since the matrix of $T$ with respect to $(v_1,...,v_n)$ is upper-triangular, this implies that for $i>j$, the $i$-coordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$ is $0$, this implies that $T(v_j)in span(v_1,...,v_j)$.
$c$ implies $b$ if $Span(v_1,...,v_j)$ is invariant by $T$, it implies that $T(v_j)in span(v_1,...,v_j)$ since $v_jin (v_1,...,v_j)$
answered Dec 16 '18 at 8:06
Tsemo AristideTsemo Aristide
59.1k11445
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add a comment |
$begingroup$
$a$ implies $b$ The $j$-column of the matrix of $T$ with respect to $v_1,...,v_n$ are the cordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$. Since the matrix of $T$ with respect to $(v_1,...,v_n)$ is upper-triangular, this implies that for $i>j$, the $i$-coordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$ is $0$, this implies that $T(v_j)in span(v_1,...,v_j)$.
$c$ implies $b$ if $Span(v_1,...,v_j)$ is invariant by $T$, it implies that $T(v_j)in span(v_1,...,v_j)$ since $v_jin (v_1,...,v_j)$
answered Dec 16 '18 at 8:06
Tsemo AristideTsemo Aristide
59.1k11445
$endgroup$
add a comment |
$begingroup$
$a$ implies $b$ The $j$-column of the matrix of $T$ with respect to $v_1,...,v_n$ are the cordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$. Since the matrix of $T$ with respect to $(v_1,...,v_n)$ is upper-triangular, this implies that for $i>j$, the $i$-coordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$ is $0$, this implies that $T(v_j)in span(v_1,...,v_j)$.
$c$ implies $b$ if $Span(v_1,...,v_j)$ is invariant by $T$, it implies that $T(v_j)in span(v_1,...,v_j)$ since $v_jin (v_1,...,v_j)$
answered Dec 16 '18 at 8:06
Tsemo AristideTsemo Aristide
59.1k11445
$endgroup$
$a$ implies $b$ The $j$-column of the matrix of $T$ with respect to $v_1,...,v_n$ are the cordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$. Since the matrix of $T$ with respect to $(v_1,...,v_n)$ is upper-triangular, this implies that for $i>j$, the $i$-coordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$ is $0$, this implies that $T(v_j)in span(v_1,...,v_j)$.
$c$ implies $b$ if $Span(v_1,...,v_j)$ is invariant by $T$, it implies that $T(v_j)in span(v_1,...,v_j)$ since $v_jin (v_1,...,v_j)$
answered Dec 16 '18 at 8:06
Tsemo AristideTsemo Aristide
59.1k11445
answered Dec 16 '18 at 8:06
Tsemo AristideTsemo Aristide
59.1k11445
answered Dec 16 '18 at 8:06
Tsemo AristideTsemo Aristide
59.1k11445
answered Dec 16 '18 at 8:06
Tsemo AristideTsemo Aristide
59.1k11445
59.1k11445
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