Proof: A Triangular Matrix is Invertible $ Longleftrightarrow $ its Eigenvalues are Real and Nonzero
$begingroup$
Problem
Prove that a triangular matrix is invertible iff its eigenvalues are real and nonzero.
Attempt
Let's call this triangular matrix $A$.
From intuition (from the invertability of A), I quickly noted that:
$$ Avec{x} = lambda Ivec{x} $$
$$ A^{-1}Avec{x} = lambda I A^{-1}Ivec{x} $$
$$ vec{x} = lambda IA^{-1}vec{x} $$
$$ frac{1}{lambda}I vec{x} = A^{-1}vec{x} $$
$$ A^{-1}vec{x} = frac{1}{lambda}I vec{x} $$
So, the eigenvalues of $ A^{-1} $ are the reciprocals of the eigenvalues of $ A $. However, not sure how to proceed from here.
Thanks in advance!
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 2 '18 at 4:42
Anthony KrivonosAnthony Krivonos
20710
$endgroup$
add a comment |
$begingroup$
Problem
Prove that a triangular matrix is invertible iff its eigenvalues are real and nonzero.
Attempt
Let's call this triangular matrix $A$.
From intuition (from the invertability of A), I quickly noted that:
$$ Avec{x} = lambda Ivec{x} $$
$$ A^{-1}Avec{x} = lambda I A^{-1}Ivec{x} $$
$$ vec{x} = lambda IA^{-1}vec{x} $$
$$ frac{1}{lambda}I vec{x} = A^{-1}vec{x} $$
$$ A^{-1}vec{x} = frac{1}{lambda}I vec{x} $$
So, the eigenvalues of $ A^{-1} $ are the reciprocals of the eigenvalues of $ A $. However, not sure how to proceed from here.
Thanks in advance!
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 2 '18 at 4:42
Anthony KrivonosAnthony Krivonos
20710
$endgroup$
add a comment |
$begingroup$
Problem
Prove that a triangular matrix is invertible iff its eigenvalues are real and nonzero.
Attempt
Let's call this triangular matrix $A$.
From intuition (from the invertability of A), I quickly noted that:
$$ Avec{x} = lambda Ivec{x} $$
$$ A^{-1}Avec{x} = lambda I A^{-1}Ivec{x} $$
$$ vec{x} = lambda IA^{-1}vec{x} $$
$$ frac{1}{lambda}I vec{x} = A^{-1}vec{x} $$
$$ A^{-1}vec{x} = frac{1}{lambda}I vec{x} $$
So, the eigenvalues of $ A^{-1} $ are the reciprocals of the eigenvalues of $ A $. However, not sure how to proceed from here.
Thanks in advance!
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 2 '18 at 4:42
Anthony KrivonosAnthony Krivonos
20710
$endgroup$
Problem
Prove that a triangular matrix is invertible iff its eigenvalues are real and nonzero.
Attempt
Let's call this triangular matrix $A$.
From intuition (from the invertability of A), I quickly noted that:
$$ Avec{x} = lambda Ivec{x} $$
$$ A^{-1}Avec{x} = lambda I A^{-1}Ivec{x} $$
$$ vec{x} = lambda IA^{-1}vec{x} $$
$$ frac{1}{lambda}I vec{x} = A^{-1}vec{x} $$
$$ A^{-1}vec{x} = frac{1}{lambda}I vec{x} $$
So, the eigenvalues of $ A^{-1} $ are the reciprocals of the eigenvalues of $ A $. However, not sure how to proceed from here.
Thanks in advance!
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 2 '18 at 4:42
Anthony KrivonosAnthony Krivonos
20710
asked Dec 2 '18 at 4:42
Anthony KrivonosAnthony Krivonos
20710
asked Dec 2 '18 at 4:42
Anthony KrivonosAnthony Krivonos
20710
asked Dec 2 '18 at 4:42
Anthony KrivonosAnthony Krivonos
20710
asked Dec 2 '18 at 4:42
Anthony KrivonosAnthony Krivonos
20710
20710
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I believe the assumption here is that $A$ is a real triangular matrix.
$$det(A-lambda I)=0$$
$$prod_{i=1}^n (A_{ii}-lambda)=0$$
The diagonal entries are the eigenvalues.
If $A$ is invertible, its determinant is non-zero. Hence, all the diagonal entries are non-zero and hence all the eigenvalues are non-zero.
Also, if the eigenvalues are real non-zero, then all the diagonal entries ae non-zero and hence $A$ is invertible.
answered Dec 2 '18 at 4:52
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
$begingroup$
Woah, this is clear as he$mathcal{L}$$mathcal{L}$. Thanks for the help!
$endgroup$
– Anthony Krivonos
Dec 2 '18 at 5:00
add a comment |
$begingroup$
Notice that for a triangular matrix eigen values are nothing but diagonal entries and determinant of a tiangular matrix is product of diagonal enties. Note another thing a matrix is invertible iff it's determinant is non zero.
It is enough to prove the second statement as for a triangular matrix $A$ the matrix $A-lambda I$ is also tiangular. Where $I$ is the identity matrix of order same as order of $A$ and $lambda$ is an indeterminate.
Notice the formula of determinant $$det(B)=sum_{sigma in S_n} b_{1,sigma(1)}.....b_{n,sigma(n)}$$ ,where $B$ is a $n×n$ matrix and $S_n$ is the symmetric group of ${1,2,....,n}$. Notice that whenever $B$ is triangular then each term of summation is zero ,except one term ( this particular term is nothing but product of diagonal entries), as each term of summation contains exactly one element from each column and exactly one element from each row.
answered Dec 2 '18 at 5:00
UserSUserS
1,5391112
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I believe the assumption here is that $A$ is a real triangular matrix.
$$det(A-lambda I)=0$$
$$prod_{i=1}^n (A_{ii}-lambda)=0$$
The diagonal entries are the eigenvalues.
If $A$ is invertible, its determinant is non-zero. Hence, all the diagonal entries are non-zero and hence all the eigenvalues are non-zero.
Also, if the eigenvalues are real non-zero, then all the diagonal entries ae non-zero and hence $A$ is invertible.
answered Dec 2 '18 at 4:52
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
$begingroup$
Woah, this is clear as he$mathcal{L}$$mathcal{L}$. Thanks for the help!
$endgroup$
– Anthony Krivonos
Dec 2 '18 at 5:00
add a comment |
$begingroup$
I believe the assumption here is that $A$ is a real triangular matrix.
$$det(A-lambda I)=0$$
$$prod_{i=1}^n (A_{ii}-lambda)=0$$
The diagonal entries are the eigenvalues.
If $A$ is invertible, its determinant is non-zero. Hence, all the diagonal entries are non-zero and hence all the eigenvalues are non-zero.
Also, if the eigenvalues are real non-zero, then all the diagonal entries ae non-zero and hence $A$ is invertible.
answered Dec 2 '18 at 4:52
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
$begingroup$
Woah, this is clear as he$mathcal{L}$$mathcal{L}$. Thanks for the help!
$endgroup$
– Anthony Krivonos
Dec 2 '18 at 5:00
add a comment |
$begingroup$
I believe the assumption here is that $A$ is a real triangular matrix.
$$det(A-lambda I)=0$$
$$prod_{i=1}^n (A_{ii}-lambda)=0$$
The diagonal entries are the eigenvalues.
If $A$ is invertible, its determinant is non-zero. Hence, all the diagonal entries are non-zero and hence all the eigenvalues are non-zero.
Also, if the eigenvalues are real non-zero, then all the diagonal entries ae non-zero and hence $A$ is invertible.
answered Dec 2 '18 at 4:52
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
I believe the assumption here is that $A$ is a real triangular matrix.
$$det(A-lambda I)=0$$
$$prod_{i=1}^n (A_{ii}-lambda)=0$$
The diagonal entries are the eigenvalues.
If $A$ is invertible, its determinant is non-zero. Hence, all the diagonal entries are non-zero and hence all the eigenvalues are non-zero.
Also, if the eigenvalues are real non-zero, then all the diagonal entries ae non-zero and hence $A$ is invertible.
answered Dec 2 '18 at 4:52
Siong Thye GohSiong Thye Goh
100k1465117
answered Dec 2 '18 at 4:52
Siong Thye GohSiong Thye Goh
100k1465117
answered Dec 2 '18 at 4:52
Siong Thye GohSiong Thye Goh
100k1465117
answered Dec 2 '18 at 4:52
Siong Thye GohSiong Thye Goh
100k1465117
100k1465117
$begingroup$
Woah, this is clear as he$mathcal{L}$$mathcal{L}$. Thanks for the help!
$endgroup$
– Anthony Krivonos
Dec 2 '18 at 5:00
add a comment |
$begingroup$
Woah, this is clear as he$mathcal{L}$$mathcal{L}$. Thanks for the help!
$endgroup$
– Anthony Krivonos
Dec 2 '18 at 5:00
$begingroup$
Woah, this is clear as he$mathcal{L}$$mathcal{L}$. Thanks for the help!
$endgroup$
– Anthony Krivonos
Dec 2 '18 at 5:00
$begingroup$
Woah, this is clear as he$mathcal{L}$$mathcal{L}$. Thanks for the help!
$endgroup$
– Anthony Krivonos
Dec 2 '18 at 5:00
add a comment |
$begingroup$
Notice that for a triangular matrix eigen values are nothing but diagonal entries and determinant of a tiangular matrix is product of diagonal enties. Note another thing a matrix is invertible iff it's determinant is non zero.
It is enough to prove the second statement as for a triangular matrix $A$ the matrix $A-lambda I$ is also tiangular. Where $I$ is the identity matrix of order same as order of $A$ and $lambda$ is an indeterminate.
Notice the formula of determinant $$det(B)=sum_{sigma in S_n} b_{1,sigma(1)}.....b_{n,sigma(n)}$$ ,where $B$ is a $n×n$ matrix and $S_n$ is the symmetric group of ${1,2,....,n}$. Notice that whenever $B$ is triangular then each term of summation is zero ,except one term ( this particular term is nothing but product of diagonal entries), as each term of summation contains exactly one element from each column and exactly one element from each row.
answered Dec 2 '18 at 5:00
UserSUserS
1,5391112
$endgroup$
add a comment |
$begingroup$
Notice that for a triangular matrix eigen values are nothing but diagonal entries and determinant of a tiangular matrix is product of diagonal enties. Note another thing a matrix is invertible iff it's determinant is non zero.
It is enough to prove the second statement as for a triangular matrix $A$ the matrix $A-lambda I$ is also tiangular. Where $I$ is the identity matrix of order same as order of $A$ and $lambda$ is an indeterminate.
Notice the formula of determinant $$det(B)=sum_{sigma in S_n} b_{1,sigma(1)}.....b_{n,sigma(n)}$$ ,where $B$ is a $n×n$ matrix and $S_n$ is the symmetric group of ${1,2,....,n}$. Notice that whenever $B$ is triangular then each term of summation is zero ,except one term ( this particular term is nothing but product of diagonal entries), as each term of summation contains exactly one element from each column and exactly one element from each row.
answered Dec 2 '18 at 5:00
UserSUserS
1,5391112
$endgroup$
add a comment |
$begingroup$
Notice that for a triangular matrix eigen values are nothing but diagonal entries and determinant of a tiangular matrix is product of diagonal enties. Note another thing a matrix is invertible iff it's determinant is non zero.
It is enough to prove the second statement as for a triangular matrix $A$ the matrix $A-lambda I$ is also tiangular. Where $I$ is the identity matrix of order same as order of $A$ and $lambda$ is an indeterminate.
Notice the formula of determinant $$det(B)=sum_{sigma in S_n} b_{1,sigma(1)}.....b_{n,sigma(n)}$$ ,where $B$ is a $n×n$ matrix and $S_n$ is the symmetric group of ${1,2,....,n}$. Notice that whenever $B$ is triangular then each term of summation is zero ,except one term ( this particular term is nothing but product of diagonal entries), as each term of summation contains exactly one element from each column and exactly one element from each row.
answered Dec 2 '18 at 5:00
UserSUserS
1,5391112
$endgroup$
Notice that for a triangular matrix eigen values are nothing but diagonal entries and determinant of a tiangular matrix is product of diagonal enties. Note another thing a matrix is invertible iff it's determinant is non zero.
It is enough to prove the second statement as for a triangular matrix $A$ the matrix $A-lambda I$ is also tiangular. Where $I$ is the identity matrix of order same as order of $A$ and $lambda$ is an indeterminate.
Notice the formula of determinant $$det(B)=sum_{sigma in S_n} b_{1,sigma(1)}.....b_{n,sigma(n)}$$ ,where $B$ is a $n×n$ matrix and $S_n$ is the symmetric group of ${1,2,....,n}$. Notice that whenever $B$ is triangular then each term of summation is zero ,except one term ( this particular term is nothing but product of diagonal entries), as each term of summation contains exactly one element from each column and exactly one element from each row.
answered Dec 2 '18 at 5:00
UserSUserS
1,5391112
edited Dec 2 '18 at 5:07
answered Dec 2 '18 at 5:00
UserSUserS
1,5391112
answered Dec 2 '18 at 5:00
UserSUserS
1,5391112
answered Dec 2 '18 at 5:00
UserSUserS
1,5391112
1,5391112
add a comment |
add a comment |
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