Invertible Matrices within a Matrix
$begingroup$
Suppose A, B are invertible matrices of the same size. Show that
$$M = begin{bmatrix} 0& A\ B& 0end{bmatrix}$$ is invertible.
I don't understand how I could show this. I have learned about linear combinations and spanning in my college class, but I don't know how that would help in this case.
linear-algebra matrices inverse block-matrices
edited Jan 31 at 5:44
Martin Sleziak
44.7k10119272
asked Jan 31 at 4:33
BaileyBailey
361
$endgroup$
add a comment |
$begingroup$
Suppose A, B are invertible matrices of the same size. Show that
$$M = begin{bmatrix} 0& A\ B& 0end{bmatrix}$$ is invertible.
I don't understand how I could show this. I have learned about linear combinations and spanning in my college class, but I don't know how that would help in this case.
linear-algebra matrices inverse block-matrices
edited Jan 31 at 5:44
Martin Sleziak
44.7k10119272
asked Jan 31 at 4:33
BaileyBailey
361
$endgroup$
add a comment |
$begingroup$
Suppose A, B are invertible matrices of the same size. Show that
$$M = begin{bmatrix} 0& A\ B& 0end{bmatrix}$$ is invertible.
I don't understand how I could show this. I have learned about linear combinations and spanning in my college class, but I don't know how that would help in this case.
linear-algebra matrices inverse block-matrices
edited Jan 31 at 5:44
Martin Sleziak
44.7k10119272
asked Jan 31 at 4:33
BaileyBailey
361
$endgroup$
Suppose A, B are invertible matrices of the same size. Show that
$$M = begin{bmatrix} 0& A\ B& 0end{bmatrix}$$ is invertible.
I don't understand how I could show this. I have learned about linear combinations and spanning in my college class, but I don't know how that would help in this case.
linear-algebra matrices inverse block-matrices
linear-algebra matrices inverse block-matrices
edited Jan 31 at 5:44
Martin Sleziak
44.7k10119272
asked Jan 31 at 4:33
BaileyBailey
361
edited Jan 31 at 5:44
Martin Sleziak
44.7k10119272
asked Jan 31 at 4:33
BaileyBailey
361
edited Jan 31 at 5:44
Martin Sleziak
44.7k10119272
edited Jan 31 at 5:44
Martin Sleziak
44.7k10119272
edited Jan 31 at 5:44
Martin Sleziak
44.7k10119272
44.7k10119272
asked Jan 31 at 4:33
BaileyBailey
361
asked Jan 31 at 4:33
BaileyBailey
361
asked Jan 31 at 4:33
BaileyBailey
361
361
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You can check directly that
$$
M^{-1}=begin{bmatrix} 0&B^{-1}\ A^{-1}&0end{bmatrix}.
$$
answered Jan 31 at 4:40
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
$begingroup$
Note that $M begin{bmatrix} x \ y end{bmatrix} = begin{bmatrix} w \ z end{bmatrix}$ iff $A y = w$ and $Bx = z$.
From this you can compute an explicit inverse.
answered Jan 31 at 4:35
copper.hatcopper.hat
127k559160
$endgroup$
add a comment |
$begingroup$
Since $A$ and $B$ are invertible, we have
$M begin{bmatrix} x \ y end{bmatrix}=begin{bmatrix} 0 \ 0 end{bmatrix} iff Ay=0$ and $Bx=0 iff x=y=0.$
answered Jan 31 at 6:25
FredFred
46.9k1848
$endgroup$
add a comment |
$begingroup$
Since you mentioned linear combinations, you can also follow that path and prove that all the matrix rows are linearly independent, which is equivalent to invertibility.
For that, let $a_i$ and $b_i$ be the $i$-th rows of $A$ and $B$, respectively. Now assume that we have a linear combination of the rows of $M$. These rows have the form $(0|a_i)$ (some zeros followed by a row of $A$) or $(b_i|0)$ (a row of $B$ followed by some zeros). A linear combination of the rows of $M$ is therefore written as
$$
sum_i alpha_i(0|a_i) + sum_i beta_i(b_i|0)
$$
where $alpha_i$ and $beta_i$ are some scalar coefficients. Assume the sum above is the null vector, and let's prove that the coefficients must be zero.
Now since the number of added zeros in the vectors $(0|a_i)$ and $(b_i|0)$ is exactly the size of $A$ and $B$, the above sum can be written as
$$
(sum_i alpha_i a_i | sum_i beta_i b_i)
$$
This is the null vector only when we have both
$$
sum_i alpha_i a_i = 0 qquad sum_i beta_i b_i = 0
$$
Since $A$ and $B$ are invertible, their rows are linearly independent, so in the sums above we must have $alpha_i=0$ and $beta_i=0$, for all $i$. QED.
answered Jan 31 at 8:55
chichi
1,424813
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can check directly that
$$
M^{-1}=begin{bmatrix} 0&B^{-1}\ A^{-1}&0end{bmatrix}.
$$
answered Jan 31 at 4:40
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
$begingroup$
You can check directly that
$$
M^{-1}=begin{bmatrix} 0&B^{-1}\ A^{-1}&0end{bmatrix}.
$$
answered Jan 31 at 4:40
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
$begingroup$
You can check directly that
$$
M^{-1}=begin{bmatrix} 0&B^{-1}\ A^{-1}&0end{bmatrix}.
$$
answered Jan 31 at 4:40
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
You can check directly that
$$
M^{-1}=begin{bmatrix} 0&B^{-1}\ A^{-1}&0end{bmatrix}.
$$
answered Jan 31 at 4:40
Martin ArgeramiMartin Argerami
127k1182183
answered Jan 31 at 4:40
Martin ArgeramiMartin Argerami
127k1182183
answered Jan 31 at 4:40
Martin ArgeramiMartin Argerami
127k1182183
answered Jan 31 at 4:40
Martin ArgeramiMartin Argerami
127k1182183
127k1182183
add a comment |
add a comment |
$begingroup$
Note that $M begin{bmatrix} x \ y end{bmatrix} = begin{bmatrix} w \ z end{bmatrix}$ iff $A y = w$ and $Bx = z$.
From this you can compute an explicit inverse.
answered Jan 31 at 4:35
copper.hatcopper.hat
127k559160
$endgroup$
add a comment |
$begingroup$
Note that $M begin{bmatrix} x \ y end{bmatrix} = begin{bmatrix} w \ z end{bmatrix}$ iff $A y = w$ and $Bx = z$.
From this you can compute an explicit inverse.
answered Jan 31 at 4:35
copper.hatcopper.hat
127k559160
$endgroup$
add a comment |
$begingroup$
Note that $M begin{bmatrix} x \ y end{bmatrix} = begin{bmatrix} w \ z end{bmatrix}$ iff $A y = w$ and $Bx = z$.
From this you can compute an explicit inverse.
answered Jan 31 at 4:35
copper.hatcopper.hat
127k559160
$endgroup$
Note that $M begin{bmatrix} x \ y end{bmatrix} = begin{bmatrix} w \ z end{bmatrix}$ iff $A y = w$ and $Bx = z$.
From this you can compute an explicit inverse.
answered Jan 31 at 4:35
copper.hatcopper.hat
127k559160
answered Jan 31 at 4:35
copper.hatcopper.hat
127k559160
answered Jan 31 at 4:35
copper.hatcopper.hat
127k559160
answered Jan 31 at 4:35
copper.hatcopper.hat
127k559160
127k559160
add a comment |
add a comment |
$begingroup$
Since $A$ and $B$ are invertible, we have
$M begin{bmatrix} x \ y end{bmatrix}=begin{bmatrix} 0 \ 0 end{bmatrix} iff Ay=0$ and $Bx=0 iff x=y=0.$
answered Jan 31 at 6:25
FredFred
46.9k1848
$endgroup$
add a comment |
$begingroup$
Since $A$ and $B$ are invertible, we have
$M begin{bmatrix} x \ y end{bmatrix}=begin{bmatrix} 0 \ 0 end{bmatrix} iff Ay=0$ and $Bx=0 iff x=y=0.$
answered Jan 31 at 6:25
FredFred
46.9k1848
$endgroup$
add a comment |
$begingroup$
Since $A$ and $B$ are invertible, we have
$M begin{bmatrix} x \ y end{bmatrix}=begin{bmatrix} 0 \ 0 end{bmatrix} iff Ay=0$ and $Bx=0 iff x=y=0.$
answered Jan 31 at 6:25
FredFred
46.9k1848
$endgroup$
Since $A$ and $B$ are invertible, we have
$M begin{bmatrix} x \ y end{bmatrix}=begin{bmatrix} 0 \ 0 end{bmatrix} iff Ay=0$ and $Bx=0 iff x=y=0.$
answered Jan 31 at 6:25
FredFred
46.9k1848
answered Jan 31 at 6:25
FredFred
46.9k1848
answered Jan 31 at 6:25
FredFred
46.9k1848
answered Jan 31 at 6:25
FredFred
46.9k1848
46.9k1848
add a comment |
add a comment |
$begingroup$
Since you mentioned linear combinations, you can also follow that path and prove that all the matrix rows are linearly independent, which is equivalent to invertibility.
For that, let $a_i$ and $b_i$ be the $i$-th rows of $A$ and $B$, respectively. Now assume that we have a linear combination of the rows of $M$. These rows have the form $(0|a_i)$ (some zeros followed by a row of $A$) or $(b_i|0)$ (a row of $B$ followed by some zeros). A linear combination of the rows of $M$ is therefore written as
$$
sum_i alpha_i(0|a_i) + sum_i beta_i(b_i|0)
$$
where $alpha_i$ and $beta_i$ are some scalar coefficients. Assume the sum above is the null vector, and let's prove that the coefficients must be zero.
Now since the number of added zeros in the vectors $(0|a_i)$ and $(b_i|0)$ is exactly the size of $A$ and $B$, the above sum can be written as
$$
(sum_i alpha_i a_i | sum_i beta_i b_i)
$$
This is the null vector only when we have both
$$
sum_i alpha_i a_i = 0 qquad sum_i beta_i b_i = 0
$$
Since $A$ and $B$ are invertible, their rows are linearly independent, so in the sums above we must have $alpha_i=0$ and $beta_i=0$, for all $i$. QED.
answered Jan 31 at 8:55
chichi
1,424813
$endgroup$
add a comment |
$begingroup$
Since you mentioned linear combinations, you can also follow that path and prove that all the matrix rows are linearly independent, which is equivalent to invertibility.
For that, let $a_i$ and $b_i$ be the $i$-th rows of $A$ and $B$, respectively. Now assume that we have a linear combination of the rows of $M$. These rows have the form $(0|a_i)$ (some zeros followed by a row of $A$) or $(b_i|0)$ (a row of $B$ followed by some zeros). A linear combination of the rows of $M$ is therefore written as
$$
sum_i alpha_i(0|a_i) + sum_i beta_i(b_i|0)
$$
where $alpha_i$ and $beta_i$ are some scalar coefficients. Assume the sum above is the null vector, and let's prove that the coefficients must be zero.
Now since the number of added zeros in the vectors $(0|a_i)$ and $(b_i|0)$ is exactly the size of $A$ and $B$, the above sum can be written as
$$
(sum_i alpha_i a_i | sum_i beta_i b_i)
$$
This is the null vector only when we have both
$$
sum_i alpha_i a_i = 0 qquad sum_i beta_i b_i = 0
$$
Since $A$ and $B$ are invertible, their rows are linearly independent, so in the sums above we must have $alpha_i=0$ and $beta_i=0$, for all $i$. QED.
answered Jan 31 at 8:55
chichi
1,424813
$endgroup$
add a comment |
$begingroup$
Since you mentioned linear combinations, you can also follow that path and prove that all the matrix rows are linearly independent, which is equivalent to invertibility.
For that, let $a_i$ and $b_i$ be the $i$-th rows of $A$ and $B$, respectively. Now assume that we have a linear combination of the rows of $M$. These rows have the form $(0|a_i)$ (some zeros followed by a row of $A$) or $(b_i|0)$ (a row of $B$ followed by some zeros). A linear combination of the rows of $M$ is therefore written as
$$
sum_i alpha_i(0|a_i) + sum_i beta_i(b_i|0)
$$
where $alpha_i$ and $beta_i$ are some scalar coefficients. Assume the sum above is the null vector, and let's prove that the coefficients must be zero.
Now since the number of added zeros in the vectors $(0|a_i)$ and $(b_i|0)$ is exactly the size of $A$ and $B$, the above sum can be written as
$$
(sum_i alpha_i a_i | sum_i beta_i b_i)
$$
This is the null vector only when we have both
$$
sum_i alpha_i a_i = 0 qquad sum_i beta_i b_i = 0
$$
Since $A$ and $B$ are invertible, their rows are linearly independent, so in the sums above we must have $alpha_i=0$ and $beta_i=0$, for all $i$. QED.
answered Jan 31 at 8:55
chichi
1,424813
$endgroup$
Since you mentioned linear combinations, you can also follow that path and prove that all the matrix rows are linearly independent, which is equivalent to invertibility.
For that, let $a_i$ and $b_i$ be the $i$-th rows of $A$ and $B$, respectively. Now assume that we have a linear combination of the rows of $M$. These rows have the form $(0|a_i)$ (some zeros followed by a row of $A$) or $(b_i|0)$ (a row of $B$ followed by some zeros). A linear combination of the rows of $M$ is therefore written as
$$
sum_i alpha_i(0|a_i) + sum_i beta_i(b_i|0)
$$
where $alpha_i$ and $beta_i$ are some scalar coefficients. Assume the sum above is the null vector, and let's prove that the coefficients must be zero.
Now since the number of added zeros in the vectors $(0|a_i)$ and $(b_i|0)$ is exactly the size of $A$ and $B$, the above sum can be written as
$$
(sum_i alpha_i a_i | sum_i beta_i b_i)
$$
This is the null vector only when we have both
$$
sum_i alpha_i a_i = 0 qquad sum_i beta_i b_i = 0
$$
Since $A$ and $B$ are invertible, their rows are linearly independent, so in the sums above we must have $alpha_i=0$ and $beta_i=0$, for all $i$. QED.
answered Jan 31 at 8:55
chichi
1,424813
answered Jan 31 at 8:55
chichi
1,424813
answered Jan 31 at 8:55
chichi
1,424813
answered Jan 31 at 8:55
chichi
1,424813
1,424813
add a comment |
add a comment |
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