Mixing
Equivalence of a scalar to a vector:
$begingroup$
I just got myself into this position while studying matrix operation, I am sure others must have been there before . If A is a matrix of dimension (m,n) and k is a scalar then ( A + k ) gives the element wise addition of the scalar k to A.
So, A + k = A + K -----(1),
where K is a matrix /vector of dimension (m,n) and have entries all equal to k.
Now can I say , from (1) that , scalar k = K , a vector of dimension (m,n)?
linear-algebra
asked Jan 20 at 9:59
shadow khshadow kh
543211
$endgroup$
add a comment |
$begingroup$
I just got myself into this position while studying matrix operation, I am sure others must have been there before . If A is a matrix of dimension (m,n) and k is a scalar then ( A + k ) gives the element wise addition of the scalar k to A.
So, A + k = A + K -----(1),
where K is a matrix /vector of dimension (m,n) and have entries all equal to k.
Now can I say , from (1) that , scalar k = K , a vector of dimension (m,n)?
linear-algebra
asked Jan 20 at 9:59
shadow khshadow kh
543211
$endgroup$
1
$begingroup$
All of these operations look highly unusual. You cannot add (m,n) and (m,1), the dimensions don't match. Even if you want an element-wise addition of a "scalar", this would make the scalar eqivalent to (m,n) matrix, not (m,1). Also, "k" is not really a scalar in this case. Scalars scale vectors if applied to them; they are only defined in (n,n) case (automorphisms) and are equivalent to a multiple of a unit square matrix, not a matrix of all equal elements.
$endgroup$
– orion
Jan 20 at 10:15
1
$begingroup$
No. It's just a notation, and doesn't make a scalar be equal to a matrix. Btw, $K$ should be also an $mtimes n$ matrix.
$endgroup$
– Berci
Jan 20 at 10:16
$begingroup$
ok if K is equal to an m×n matrix , then from (1) , K = k ?
$endgroup$
– shadow kh
Jan 20 at 10:19
add a comment |
$begingroup$
I just got myself into this position while studying matrix operation, I am sure others must have been there before . If A is a matrix of dimension (m,n) and k is a scalar then ( A + k ) gives the element wise addition of the scalar k to A.
So, A + k = A + K -----(1),
where K is a matrix /vector of dimension (m,n) and have entries all equal to k.
Now can I say , from (1) that , scalar k = K , a vector of dimension (m,n)?
linear-algebra
asked Jan 20 at 9:59
shadow khshadow kh
543211
$endgroup$
I just got myself into this position while studying matrix operation, I am sure others must have been there before . If A is a matrix of dimension (m,n) and k is a scalar then ( A + k ) gives the element wise addition of the scalar k to A.
So, A + k = A + K -----(1),
where K is a matrix /vector of dimension (m,n) and have entries all equal to k.
Now can I say , from (1) that , scalar k = K , a vector of dimension (m,n)?
linear-algebra
linear-algebra
asked Jan 20 at 9:59
shadow khshadow kh
543211
asked Jan 20 at 9:59
shadow khshadow kh
543211
edited Jan 20 at 10:20
shadow kh
asked Jan 20 at 9:59
shadow khshadow kh
543211
asked Jan 20 at 9:59
shadow khshadow kh
543211
asked Jan 20 at 9:59
shadow khshadow kh
543211
543211
1
$begingroup$
All of these operations look highly unusual. You cannot add (m,n) and (m,1), the dimensions don't match. Even if you want an element-wise addition of a "scalar", this would make the scalar eqivalent to (m,n) matrix, not (m,1). Also, "k" is not really a scalar in this case. Scalars scale vectors if applied to them; they are only defined in (n,n) case (automorphisms) and are equivalent to a multiple of a unit square matrix, not a matrix of all equal elements.
$endgroup$
– orion
Jan 20 at 10:15
1
$begingroup$
No. It's just a notation, and doesn't make a scalar be equal to a matrix. Btw, $K$ should be also an $mtimes n$ matrix.
$endgroup$
– Berci
Jan 20 at 10:16
$begingroup$
ok if K is equal to an m×n matrix , then from (1) , K = k ?
$endgroup$
– shadow kh
Jan 20 at 10:19
add a comment |
1
$begingroup$
All of these operations look highly unusual. You cannot add (m,n) and (m,1), the dimensions don't match. Even if you want an element-wise addition of a "scalar", this would make the scalar eqivalent to (m,n) matrix, not (m,1). Also, "k" is not really a scalar in this case. Scalars scale vectors if applied to them; they are only defined in (n,n) case (automorphisms) and are equivalent to a multiple of a unit square matrix, not a matrix of all equal elements.
$endgroup$
– orion
Jan 20 at 10:15
1
$begingroup$
No. It's just a notation, and doesn't make a scalar be equal to a matrix. Btw, $K$ should be also an $mtimes n$ matrix.
$endgroup$
– Berci
Jan 20 at 10:16
$begingroup$
ok if K is equal to an m×n matrix , then from (1) , K = k ?
$endgroup$
– shadow kh
Jan 20 at 10:19
1
1
$begingroup$
All of these operations look highly unusual. You cannot add (m,n) and (m,1), the dimensions don't match. Even if you want an element-wise addition of a "scalar", this would make the scalar eqivalent to (m,n) matrix, not (m,1). Also, "k" is not really a scalar in this case. Scalars scale vectors if applied to them; they are only defined in (n,n) case (automorphisms) and are equivalent to a multiple of a unit square matrix, not a matrix of all equal elements.
$endgroup$
– orion
Jan 20 at 10:15
$begingroup$
All of these operations look highly unusual. You cannot add (m,n) and (m,1), the dimensions don't match. Even if you want an element-wise addition of a "scalar", this would make the scalar eqivalent to (m,n) matrix, not (m,1). Also, "k" is not really a scalar in this case. Scalars scale vectors if applied to them; they are only defined in (n,n) case (automorphisms) and are equivalent to a multiple of a unit square matrix, not a matrix of all equal elements.
$endgroup$
– orion
Jan 20 at 10:15
1
1
$begingroup$
No. It's just a notation, and doesn't make a scalar be equal to a matrix. Btw, $K$ should be also an $mtimes n$ matrix.
$endgroup$
– Berci
Jan 20 at 10:16
$begingroup$
No. It's just a notation, and doesn't make a scalar be equal to a matrix. Btw, $K$ should be also an $mtimes n$ matrix.
$endgroup$
– Berci
Jan 20 at 10:16
$begingroup$
ok if K is equal to an m×n matrix , then from (1) , K = k ?
$endgroup$
– shadow kh
Jan 20 at 10:19
$begingroup$
ok if K is equal to an m×n matrix , then from (1) , K = k ?
$endgroup$
– shadow kh
Jan 20 at 10:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is definitely not an equivalence, addition of a scalar and a matrix is not defined. Although it is not commonly used, you can surely say that if $A$ is a $mtimes n$ matrix and $k$ is a scalar then $A+k$ is a shorthand for $A+K$ where $K$ is the $mtimes n$ matrix with all entries equal to $k$. But $k$ and $K$ are not the same thing, it is just notation.
A notation which is much more widely used is, if $A$ is a square matrix and $k$ a scalar, to write $A +k$ for $A+k I$, where $I$ is the identity matrix of the appropriate dimension. While more used, this is again notation, $k$ and $kI$ are not the same thing.
answered Jan 20 at 10:28
GFRGFR
3,2741023
$endgroup$
$begingroup$
u cleared my doubt.Thanks...
$endgroup$
– shadow kh
Jan 20 at 10:30
add a comment |
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$begingroup$
It is definitely not an equivalence, addition of a scalar and a matrix is not defined. Although it is not commonly used, you can surely say that if $A$ is a $mtimes n$ matrix and $k$ is a scalar then $A+k$ is a shorthand for $A+K$ where $K$ is the $mtimes n$ matrix with all entries equal to $k$. But $k$ and $K$ are not the same thing, it is just notation.
A notation which is much more widely used is, if $A$ is a square matrix and $k$ a scalar, to write $A +k$ for $A+k I$, where $I$ is the identity matrix of the appropriate dimension. While more used, this is again notation, $k$ and $kI$ are not the same thing.
answered Jan 20 at 10:28
GFRGFR
3,2741023
$endgroup$
$begingroup$
u cleared my doubt.Thanks...
$endgroup$
– shadow kh
Jan 20 at 10:30
add a comment |
$begingroup$
It is definitely not an equivalence, addition of a scalar and a matrix is not defined. Although it is not commonly used, you can surely say that if $A$ is a $mtimes n$ matrix and $k$ is a scalar then $A+k$ is a shorthand for $A+K$ where $K$ is the $mtimes n$ matrix with all entries equal to $k$. But $k$ and $K$ are not the same thing, it is just notation.
A notation which is much more widely used is, if $A$ is a square matrix and $k$ a scalar, to write $A +k$ for $A+k I$, where $I$ is the identity matrix of the appropriate dimension. While more used, this is again notation, $k$ and $kI$ are not the same thing.
answered Jan 20 at 10:28
GFRGFR
3,2741023
$endgroup$
$begingroup$
u cleared my doubt.Thanks...
$endgroup$
– shadow kh
Jan 20 at 10:30
add a comment |
$begingroup$
It is definitely not an equivalence, addition of a scalar and a matrix is not defined. Although it is not commonly used, you can surely say that if $A$ is a $mtimes n$ matrix and $k$ is a scalar then $A+k$ is a shorthand for $A+K$ where $K$ is the $mtimes n$ matrix with all entries equal to $k$. But $k$ and $K$ are not the same thing, it is just notation.
A notation which is much more widely used is, if $A$ is a square matrix and $k$ a scalar, to write $A +k$ for $A+k I$, where $I$ is the identity matrix of the appropriate dimension. While more used, this is again notation, $k$ and $kI$ are not the same thing.
answered Jan 20 at 10:28
GFRGFR
3,2741023
$endgroup$
It is definitely not an equivalence, addition of a scalar and a matrix is not defined. Although it is not commonly used, you can surely say that if $A$ is a $mtimes n$ matrix and $k$ is a scalar then $A+k$ is a shorthand for $A+K$ where $K$ is the $mtimes n$ matrix with all entries equal to $k$. But $k$ and $K$ are not the same thing, it is just notation.
A notation which is much more widely used is, if $A$ is a square matrix and $k$ a scalar, to write $A +k$ for $A+k I$, where $I$ is the identity matrix of the appropriate dimension. While more used, this is again notation, $k$ and $kI$ are not the same thing.
answered Jan 20 at 10:28
GFRGFR
3,2741023
answered Jan 20 at 10:28
GFRGFR
3,2741023
answered Jan 20 at 10:28
GFRGFR
3,2741023
answered Jan 20 at 10:28
GFRGFR
3,2741023
3,2741023
$begingroup$
u cleared my doubt.Thanks...
$endgroup$
– shadow kh
Jan 20 at 10:30
add a comment |
$begingroup$
u cleared my doubt.Thanks...
$endgroup$
– shadow kh
Jan 20 at 10:30
$begingroup$
u cleared my doubt.Thanks...
$endgroup$
– shadow kh
Jan 20 at 10:30
$begingroup$
u cleared my doubt.Thanks...
$endgroup$
– shadow kh
Jan 20 at 10:30
add a comment |
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1
$begingroup$
All of these operations look highly unusual. You cannot add (m,n) and (m,1), the dimensions don't match. Even if you want an element-wise addition of a "scalar", this would make the scalar eqivalent to (m,n) matrix, not (m,1). Also, "k" is not really a scalar in this case. Scalars scale vectors if applied to them; they are only defined in (n,n) case (automorphisms) and are equivalent to a multiple of a unit square matrix, not a matrix of all equal elements.
$endgroup$
– orion
Jan 20 at 10:15
1
$begingroup$
No. It's just a notation, and doesn't make a scalar be equal to a matrix. Btw, $K$ should be also an $mtimes n$ matrix.
$endgroup$
– Berci
Jan 20 at 10:16
$begingroup$
ok if K is equal to an m×n matrix , then from (1) , K = k ?
$endgroup$
– shadow kh
Jan 20 at 10:19