Matrices: $(X^{−1} + Y^{−1})^{−1} $
$(X^{−1} + Y^{−1})^{−1} = Y −Y(X + Y )^{−1}Y$
Here X, Y, and X + Y are nonsingular.
The only I don't get is the left part, because you cannot get rid of the whole expression by taking $X + Y$ inverse because the X and Y in the expression are also inversed. But you can also not get rid of the x and y separate because they are in the $X+Y$ expression.
Second thing aren't $X$ and $Y$ always nonsingular if $X+Y$ is nonsingular?
linear-algebra matrices
add a comment |
$(X^{−1} + Y^{−1})^{−1} = Y −Y(X + Y )^{−1}Y$
Here X, Y, and X + Y are nonsingular.
The only I don't get is the left part, because you cannot get rid of the whole expression by taking $X + Y$ inverse because the X and Y in the expression are also inversed. But you can also not get rid of the x and y separate because they are in the $X+Y$ expression.
Second thing aren't $X$ and $Y$ always nonsingular if $X+Y$ is nonsingular?
linear-algebra matrices
add a comment |
$(X^{−1} + Y^{−1})^{−1} = Y −Y(X + Y )^{−1}Y$
Here X, Y, and X + Y are nonsingular.
The only I don't get is the left part, because you cannot get rid of the whole expression by taking $X + Y$ inverse because the X and Y in the expression are also inversed. But you can also not get rid of the x and y separate because they are in the $X+Y$ expression.
Second thing aren't $X$ and $Y$ always nonsingular if $X+Y$ is nonsingular?
linear-algebra matrices
$(X^{−1} + Y^{−1})^{−1} = Y −Y(X + Y )^{−1}Y$
Here X, Y, and X + Y are nonsingular.
The only I don't get is the left part, because you cannot get rid of the whole expression by taking $X + Y$ inverse because the X and Y in the expression are also inversed. But you can also not get rid of the x and y separate because they are in the $X+Y$ expression.
Second thing aren't $X$ and $Y$ always nonsingular if $X+Y$ is nonsingular?
linear-algebra matrices
linear-algebra matrices
edited Nov 24 at 12:32
amWhy
191k28224439
191k28224439
asked Nov 24 at 11:45
Danielvanheuven
457
457
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add a comment |
3 Answers
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First note that if you swap $X$ and $Y$ you should get the same inverse, that's because sum is a commutative operation, so both $[Y - Y(X + Y)^{-1}Y]$ and $[X -X(X + Y)^{-1}X]$ are inverse (and obviouly $[Y - Y(X + Y)^{-1}Y] = [X - X(X + Y)^{-1}X]$)
To prove it just multiply by your r.h.s expression
begin{eqnarray}
(X^{-1} + Y^{-1}) (X^{-1} + Y^{-1})^{-1} &=& (X^{-1} + Y^{-1}) [Y - Y(X + Y)^{-1}Y] \
&=&(X^{-1} + Y^{-1})Y[1 - (X+Y)^{-1}Y] \
&=&X^{-1} X[1 - (X+Y)^{-1}X] + Y^{-1} Y[1 - (X+Y)^{-1}Y] \
&=&1 - (X + Y)^{-1}X + 1 - (X+Y)^{-1}Y \
&=& 1 - (X + Y)^{-1}(X + Y) + 1 \
&=& 1
end{eqnarray}
As for the second question, the answer is no. Imagine this case
$$
X = pmatrix{1 & 0 \ 0 & 0} ~~~ mbox{and} ~~~ Y = pmatrix{0 & 0 \ 0 & 1}
$$
neither $X$ nor $Y$ are invertible, but $X + Y$ is
add a comment |
In the proof $$(X^{-1}+Y^{-1})^{-1}=((I+X^{-1}Y)Y^{-1})^{-1}=Y(I+X^{-1}Y)^{-1}=Y(I-(Y^{-1}(X+Y))^{-1})=Y(I-(X+Y)^{-1}Y),$$the third $=$ is the least trivial step, using$$(I+X^{-1}Y)(I-(Y^{-1}(X+Y))^{-1})=I+X^{-1}Y-X^{-1}Y(I+Y^{-1}X)(I+Y^{-1}X)^{-1}=I.$$This argument assumes $X,,Y,,X^{-1}+Y^{-1}$ are all invertible. (If the first two weren't, there'd be no such thing as $X^{-1}+Y^{-1}$ to invert in the first place.) All other requirements that a specific matrix be invertible follow from these. In particular, the inverse we seek to calculate is nonexistent iff some other inverse we need along the way is nonexistent.
add a comment |
We can prove the given equation directly by premultiplying by $X^{-1}+Y^{-1}$, we get identity,
$(X^{-1}+Y^{-1})(Y-Y(X+Y)^{-1}Y) = (X^{-1}Y)+(I)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$
$=Big(IBig)+Big(X^{-1}YBig)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$
$=Big(IBig)+Big(X^{-1}Big)Big(I-(X(X+Y)^{-1})-(Y(X+Y)^{-1})Big)Big(YBig)$
$=Big(IBig)+Big(X^{-1}Big)Big(I-(X+Y)(X+Y)^{-1}Big)Big(YBig)$
$=Big(IBig)+Big(X^{-1}Big)Big(I-IBig)Big(YBig)$
$=I$
Similarly, you can postmultiply and do the same.
For your second doubt,I would say that what you are saying is the case for product of matrices and not for sum(Because the product of determinant of 2 matrices is the determinant of their product).
Hope it is helpful:)
add a comment |
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
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active
oldest
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active
oldest
votes
First note that if you swap $X$ and $Y$ you should get the same inverse, that's because sum is a commutative operation, so both $[Y - Y(X + Y)^{-1}Y]$ and $[X -X(X + Y)^{-1}X]$ are inverse (and obviouly $[Y - Y(X + Y)^{-1}Y] = [X - X(X + Y)^{-1}X]$)
To prove it just multiply by your r.h.s expression
begin{eqnarray}
(X^{-1} + Y^{-1}) (X^{-1} + Y^{-1})^{-1} &=& (X^{-1} + Y^{-1}) [Y - Y(X + Y)^{-1}Y] \
&=&(X^{-1} + Y^{-1})Y[1 - (X+Y)^{-1}Y] \
&=&X^{-1} X[1 - (X+Y)^{-1}X] + Y^{-1} Y[1 - (X+Y)^{-1}Y] \
&=&1 - (X + Y)^{-1}X + 1 - (X+Y)^{-1}Y \
&=& 1 - (X + Y)^{-1}(X + Y) + 1 \
&=& 1
end{eqnarray}
As for the second question, the answer is no. Imagine this case
$$
X = pmatrix{1 & 0 \ 0 & 0} ~~~ mbox{and} ~~~ Y = pmatrix{0 & 0 \ 0 & 1}
$$
neither $X$ nor $Y$ are invertible, but $X + Y$ is
add a comment |
First note that if you swap $X$ and $Y$ you should get the same inverse, that's because sum is a commutative operation, so both $[Y - Y(X + Y)^{-1}Y]$ and $[X -X(X + Y)^{-1}X]$ are inverse (and obviouly $[Y - Y(X + Y)^{-1}Y] = [X - X(X + Y)^{-1}X]$)
To prove it just multiply by your r.h.s expression
begin{eqnarray}
(X^{-1} + Y^{-1}) (X^{-1} + Y^{-1})^{-1} &=& (X^{-1} + Y^{-1}) [Y - Y(X + Y)^{-1}Y] \
&=&(X^{-1} + Y^{-1})Y[1 - (X+Y)^{-1}Y] \
&=&X^{-1} X[1 - (X+Y)^{-1}X] + Y^{-1} Y[1 - (X+Y)^{-1}Y] \
&=&1 - (X + Y)^{-1}X + 1 - (X+Y)^{-1}Y \
&=& 1 - (X + Y)^{-1}(X + Y) + 1 \
&=& 1
end{eqnarray}
As for the second question, the answer is no. Imagine this case
$$
X = pmatrix{1 & 0 \ 0 & 0} ~~~ mbox{and} ~~~ Y = pmatrix{0 & 0 \ 0 & 1}
$$
neither $X$ nor $Y$ are invertible, but $X + Y$ is
add a comment |
First note that if you swap $X$ and $Y$ you should get the same inverse, that's because sum is a commutative operation, so both $[Y - Y(X + Y)^{-1}Y]$ and $[X -X(X + Y)^{-1}X]$ are inverse (and obviouly $[Y - Y(X + Y)^{-1}Y] = [X - X(X + Y)^{-1}X]$)
To prove it just multiply by your r.h.s expression
begin{eqnarray}
(X^{-1} + Y^{-1}) (X^{-1} + Y^{-1})^{-1} &=& (X^{-1} + Y^{-1}) [Y - Y(X + Y)^{-1}Y] \
&=&(X^{-1} + Y^{-1})Y[1 - (X+Y)^{-1}Y] \
&=&X^{-1} X[1 - (X+Y)^{-1}X] + Y^{-1} Y[1 - (X+Y)^{-1}Y] \
&=&1 - (X + Y)^{-1}X + 1 - (X+Y)^{-1}Y \
&=& 1 - (X + Y)^{-1}(X + Y) + 1 \
&=& 1
end{eqnarray}
As for the second question, the answer is no. Imagine this case
$$
X = pmatrix{1 & 0 \ 0 & 0} ~~~ mbox{and} ~~~ Y = pmatrix{0 & 0 \ 0 & 1}
$$
neither $X$ nor $Y$ are invertible, but $X + Y$ is
First note that if you swap $X$ and $Y$ you should get the same inverse, that's because sum is a commutative operation, so both $[Y - Y(X + Y)^{-1}Y]$ and $[X -X(X + Y)^{-1}X]$ are inverse (and obviouly $[Y - Y(X + Y)^{-1}Y] = [X - X(X + Y)^{-1}X]$)
To prove it just multiply by your r.h.s expression
begin{eqnarray}
(X^{-1} + Y^{-1}) (X^{-1} + Y^{-1})^{-1} &=& (X^{-1} + Y^{-1}) [Y - Y(X + Y)^{-1}Y] \
&=&(X^{-1} + Y^{-1})Y[1 - (X+Y)^{-1}Y] \
&=&X^{-1} X[1 - (X+Y)^{-1}X] + Y^{-1} Y[1 - (X+Y)^{-1}Y] \
&=&1 - (X + Y)^{-1}X + 1 - (X+Y)^{-1}Y \
&=& 1 - (X + Y)^{-1}(X + Y) + 1 \
&=& 1
end{eqnarray}
As for the second question, the answer is no. Imagine this case
$$
X = pmatrix{1 & 0 \ 0 & 0} ~~~ mbox{and} ~~~ Y = pmatrix{0 & 0 \ 0 & 1}
$$
neither $X$ nor $Y$ are invertible, but $X + Y$ is
answered Nov 24 at 12:15
caverac
13k21028
13k21028
add a comment |
add a comment |
In the proof $$(X^{-1}+Y^{-1})^{-1}=((I+X^{-1}Y)Y^{-1})^{-1}=Y(I+X^{-1}Y)^{-1}=Y(I-(Y^{-1}(X+Y))^{-1})=Y(I-(X+Y)^{-1}Y),$$the third $=$ is the least trivial step, using$$(I+X^{-1}Y)(I-(Y^{-1}(X+Y))^{-1})=I+X^{-1}Y-X^{-1}Y(I+Y^{-1}X)(I+Y^{-1}X)^{-1}=I.$$This argument assumes $X,,Y,,X^{-1}+Y^{-1}$ are all invertible. (If the first two weren't, there'd be no such thing as $X^{-1}+Y^{-1}$ to invert in the first place.) All other requirements that a specific matrix be invertible follow from these. In particular, the inverse we seek to calculate is nonexistent iff some other inverse we need along the way is nonexistent.
add a comment |
In the proof $$(X^{-1}+Y^{-1})^{-1}=((I+X^{-1}Y)Y^{-1})^{-1}=Y(I+X^{-1}Y)^{-1}=Y(I-(Y^{-1}(X+Y))^{-1})=Y(I-(X+Y)^{-1}Y),$$the third $=$ is the least trivial step, using$$(I+X^{-1}Y)(I-(Y^{-1}(X+Y))^{-1})=I+X^{-1}Y-X^{-1}Y(I+Y^{-1}X)(I+Y^{-1}X)^{-1}=I.$$This argument assumes $X,,Y,,X^{-1}+Y^{-1}$ are all invertible. (If the first two weren't, there'd be no such thing as $X^{-1}+Y^{-1}$ to invert in the first place.) All other requirements that a specific matrix be invertible follow from these. In particular, the inverse we seek to calculate is nonexistent iff some other inverse we need along the way is nonexistent.
add a comment |
In the proof $$(X^{-1}+Y^{-1})^{-1}=((I+X^{-1}Y)Y^{-1})^{-1}=Y(I+X^{-1}Y)^{-1}=Y(I-(Y^{-1}(X+Y))^{-1})=Y(I-(X+Y)^{-1}Y),$$the third $=$ is the least trivial step, using$$(I+X^{-1}Y)(I-(Y^{-1}(X+Y))^{-1})=I+X^{-1}Y-X^{-1}Y(I+Y^{-1}X)(I+Y^{-1}X)^{-1}=I.$$This argument assumes $X,,Y,,X^{-1}+Y^{-1}$ are all invertible. (If the first two weren't, there'd be no such thing as $X^{-1}+Y^{-1}$ to invert in the first place.) All other requirements that a specific matrix be invertible follow from these. In particular, the inverse we seek to calculate is nonexistent iff some other inverse we need along the way is nonexistent.
In the proof $$(X^{-1}+Y^{-1})^{-1}=((I+X^{-1}Y)Y^{-1})^{-1}=Y(I+X^{-1}Y)^{-1}=Y(I-(Y^{-1}(X+Y))^{-1})=Y(I-(X+Y)^{-1}Y),$$the third $=$ is the least trivial step, using$$(I+X^{-1}Y)(I-(Y^{-1}(X+Y))^{-1})=I+X^{-1}Y-X^{-1}Y(I+Y^{-1}X)(I+Y^{-1}X)^{-1}=I.$$This argument assumes $X,,Y,,X^{-1}+Y^{-1}$ are all invertible. (If the first two weren't, there'd be no such thing as $X^{-1}+Y^{-1}$ to invert in the first place.) All other requirements that a specific matrix be invertible follow from these. In particular, the inverse we seek to calculate is nonexistent iff some other inverse we need along the way is nonexistent.
answered Nov 24 at 12:19
J.G.
21.7k21934
21.7k21934
add a comment |
add a comment |
We can prove the given equation directly by premultiplying by $X^{-1}+Y^{-1}$, we get identity,
$(X^{-1}+Y^{-1})(Y-Y(X+Y)^{-1}Y) = (X^{-1}Y)+(I)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$
$=Big(IBig)+Big(X^{-1}YBig)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$
$=Big(IBig)+Big(X^{-1}Big)Big(I-(X(X+Y)^{-1})-(Y(X+Y)^{-1})Big)Big(YBig)$
$=Big(IBig)+Big(X^{-1}Big)Big(I-(X+Y)(X+Y)^{-1}Big)Big(YBig)$
$=Big(IBig)+Big(X^{-1}Big)Big(I-IBig)Big(YBig)$
$=I$
Similarly, you can postmultiply and do the same.
For your second doubt,I would say that what you are saying is the case for product of matrices and not for sum(Because the product of determinant of 2 matrices is the determinant of their product).
Hope it is helpful:)
add a comment |
We can prove the given equation directly by premultiplying by $X^{-1}+Y^{-1}$, we get identity,
$(X^{-1}+Y^{-1})(Y-Y(X+Y)^{-1}Y) = (X^{-1}Y)+(I)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$
$=Big(IBig)+Big(X^{-1}YBig)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$
$=Big(IBig)+Big(X^{-1}Big)Big(I-(X(X+Y)^{-1})-(Y(X+Y)^{-1})Big)Big(YBig)$
$=Big(IBig)+Big(X^{-1}Big)Big(I-(X+Y)(X+Y)^{-1}Big)Big(YBig)$
$=Big(IBig)+Big(X^{-1}Big)Big(I-IBig)Big(YBig)$
$=I$
Similarly, you can postmultiply and do the same.
For your second doubt,I would say that what you are saying is the case for product of matrices and not for sum(Because the product of determinant of 2 matrices is the determinant of their product).
Hope it is helpful:)
add a comment |
We can prove the given equation directly by premultiplying by $X^{-1}+Y^{-1}$, we get identity,
$(X^{-1}+Y^{-1})(Y-Y(X+Y)^{-1}Y) = (X^{-1}Y)+(I)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$
$=Big(IBig)+Big(X^{-1}YBig)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$
$=Big(IBig)+Big(X^{-1}Big)Big(I-(X(X+Y)^{-1})-(Y(X+Y)^{-1})Big)Big(YBig)$
$=Big(IBig)+Big(X^{-1}Big)Big(I-(X+Y)(X+Y)^{-1}Big)Big(YBig)$
$=Big(IBig)+Big(X^{-1}Big)Big(I-IBig)Big(YBig)$
$=I$
Similarly, you can postmultiply and do the same.
For your second doubt,I would say that what you are saying is the case for product of matrices and not for sum(Because the product of determinant of 2 matrices is the determinant of their product).
Hope it is helpful:)
We can prove the given equation directly by premultiplying by $X^{-1}+Y^{-1}$, we get identity,
$(X^{-1}+Y^{-1})(Y-Y(X+Y)^{-1}Y) = (X^{-1}Y)+(I)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$
$=Big(IBig)+Big(X^{-1}YBig)-Big((X+Y)^{-1}YBig)-Big(X^{-1}Y(X+Y)^{-1}YBig)$
$=Big(IBig)+Big(X^{-1}Big)Big(I-(X(X+Y)^{-1})-(Y(X+Y)^{-1})Big)Big(YBig)$
$=Big(IBig)+Big(X^{-1}Big)Big(I-(X+Y)(X+Y)^{-1}Big)Big(YBig)$
$=Big(IBig)+Big(X^{-1}Big)Big(I-IBig)Big(YBig)$
$=I$
Similarly, you can postmultiply and do the same.
For your second doubt,I would say that what you are saying is the case for product of matrices and not for sum(Because the product of determinant of 2 matrices is the determinant of their product).
Hope it is helpful:)
edited Nov 24 at 12:22
answered Nov 24 at 12:14
Martund
1,344212
1,344212
add a comment |
add a comment |
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