Geometric meaning of left multiply with a diagonal matrix?
$begingroup$
I have
$m$ of column vectors in $x_i in mathbb{R}^{ntimes 1}$ as
$$mathbf{X} = begin{bmatrix}x_1 & x_2 & ldots x_m end{bmatrix},$$
Then I multiply with a diagonal matrix, note that each element does not have to be equal, other wise a scalar matrix is too easy,
$$mathbf{Lambda} = begin{bmatrix}lambda_1 & & \
& ddots & \
& & lambda_nend{bmatrix}$$
What is the geometric meaning of $mathbf{X^prime} = mathbf{Lambda X}$?
What is the geometric relationship between $mathbf{X}^prime$ and $mathbf{X}$?
It seems to be pretty difficult to describe its behavior. It does not preserve the anlge between column vector in $mathbf{X}$.
linear-algebra matrices linear-transformations
asked Jan 7 at 2:19
ArtificiallyIntelligenceArtificiallyIntelligence
300111
$endgroup$
add a comment |
$begingroup$
I have
$m$ of column vectors in $x_i in mathbb{R}^{ntimes 1}$ as
$$mathbf{X} = begin{bmatrix}x_1 & x_2 & ldots x_m end{bmatrix},$$
Then I multiply with a diagonal matrix, note that each element does not have to be equal, other wise a scalar matrix is too easy,
$$mathbf{Lambda} = begin{bmatrix}lambda_1 & & \
& ddots & \
& & lambda_nend{bmatrix}$$
What is the geometric meaning of $mathbf{X^prime} = mathbf{Lambda X}$?
What is the geometric relationship between $mathbf{X}^prime$ and $mathbf{X}$?
It seems to be pretty difficult to describe its behavior. It does not preserve the anlge between column vector in $mathbf{X}$.
linear-algebra matrices linear-transformations
asked Jan 7 at 2:19
ArtificiallyIntelligenceArtificiallyIntelligence
300111
$endgroup$
add a comment |
$begingroup$
I have
$m$ of column vectors in $x_i in mathbb{R}^{ntimes 1}$ as
$$mathbf{X} = begin{bmatrix}x_1 & x_2 & ldots x_m end{bmatrix},$$
Then I multiply with a diagonal matrix, note that each element does not have to be equal, other wise a scalar matrix is too easy,
$$mathbf{Lambda} = begin{bmatrix}lambda_1 & & \
& ddots & \
& & lambda_nend{bmatrix}$$
What is the geometric meaning of $mathbf{X^prime} = mathbf{Lambda X}$?
What is the geometric relationship between $mathbf{X}^prime$ and $mathbf{X}$?
It seems to be pretty difficult to describe its behavior. It does not preserve the anlge between column vector in $mathbf{X}$.
linear-algebra matrices linear-transformations
asked Jan 7 at 2:19
ArtificiallyIntelligenceArtificiallyIntelligence
300111
$endgroup$
I have
$m$ of column vectors in $x_i in mathbb{R}^{ntimes 1}$ as
$$mathbf{X} = begin{bmatrix}x_1 & x_2 & ldots x_m end{bmatrix},$$
Then I multiply with a diagonal matrix, note that each element does not have to be equal, other wise a scalar matrix is too easy,
$$mathbf{Lambda} = begin{bmatrix}lambda_1 & & \
& ddots & \
& & lambda_nend{bmatrix}$$
What is the geometric meaning of $mathbf{X^prime} = mathbf{Lambda X}$?
What is the geometric relationship between $mathbf{X}^prime$ and $mathbf{X}$?
It seems to be pretty difficult to describe its behavior. It does not preserve the anlge between column vector in $mathbf{X}$.
linear-algebra matrices linear-transformations
linear-algebra matrices linear-transformations
asked Jan 7 at 2:19
ArtificiallyIntelligenceArtificiallyIntelligence
300111
asked Jan 7 at 2:19
ArtificiallyIntelligenceArtificiallyIntelligence
300111
edited Jan 7 at 2:33
ArtificiallyIntelligence
asked Jan 7 at 2:19
ArtificiallyIntelligenceArtificiallyIntelligence
300111
asked Jan 7 at 2:19
ArtificiallyIntelligenceArtificiallyIntelligence
300111
asked Jan 7 at 2:19
ArtificiallyIntelligenceArtificiallyIntelligence
300111
300111
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
Suppose that all $lambda_i$ are non-zero. One way to think of this action is to consider the fact that
$$
Lambda X = (Lambda X Lambda^{-1}) (Lambda LambdaLambda^{-1}) = [X]_{mathcal B} [Lambda]_{mathcal B}
$$
The transition $X mapsto Lambda X Lambda^{-1}$ can be thought of geometrically as a change in basis. That is, we reinterpret $X$ as a transformation relative to the basis $mathcal B = {lambda_1 e_1, dots, lambda_n e_n}$, where $e_i$ is the $i$th standard basis vector.
Multiplying from the right by $Lambda$, as before, amounts to scaling the columns.
Or, if you prefer: multiply first, then change basis, since
$$
Lambda X = Lambda [X Lambda]Lambda^{-1}
$$
answered Jan 7 at 3:06
OmnomnomnomOmnomnomnom
128k791183
$endgroup$
add a comment |
$begingroup$
Notice that when you multiple $Lambda$ to X, the resultant matrix is $left[ lambda_1 x_1,,,,,lambda_2 x_2,,,ldots lambda_n x_nright]$. You multiply each column but the corresponding scalar.
The relationship between X$^prime$ and X is that it stretches (or contracts) each column vector.
answered Jan 7 at 2:29
Joel PereiraJoel Pereira
75919
$endgroup$
1
$begingroup$
Thank you for your reply. But isn't that a right multiplication of the diagonal matrix?
$endgroup$
– ArtificiallyIntelligence
Jan 7 at 2:32
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
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votes
$begingroup$
Suppose that all $lambda_i$ are non-zero. One way to think of this action is to consider the fact that
$$
Lambda X = (Lambda X Lambda^{-1}) (Lambda LambdaLambda^{-1}) = [X]_{mathcal B} [Lambda]_{mathcal B}
$$
The transition $X mapsto Lambda X Lambda^{-1}$ can be thought of geometrically as a change in basis. That is, we reinterpret $X$ as a transformation relative to the basis $mathcal B = {lambda_1 e_1, dots, lambda_n e_n}$, where $e_i$ is the $i$th standard basis vector.
Multiplying from the right by $Lambda$, as before, amounts to scaling the columns.
Or, if you prefer: multiply first, then change basis, since
$$
Lambda X = Lambda [X Lambda]Lambda^{-1}
$$
answered Jan 7 at 3:06
OmnomnomnomOmnomnomnom
128k791183
$endgroup$
add a comment |
$begingroup$
Suppose that all $lambda_i$ are non-zero. One way to think of this action is to consider the fact that
$$
Lambda X = (Lambda X Lambda^{-1}) (Lambda LambdaLambda^{-1}) = [X]_{mathcal B} [Lambda]_{mathcal B}
$$
The transition $X mapsto Lambda X Lambda^{-1}$ can be thought of geometrically as a change in basis. That is, we reinterpret $X$ as a transformation relative to the basis $mathcal B = {lambda_1 e_1, dots, lambda_n e_n}$, where $e_i$ is the $i$th standard basis vector.
Multiplying from the right by $Lambda$, as before, amounts to scaling the columns.
Or, if you prefer: multiply first, then change basis, since
$$
Lambda X = Lambda [X Lambda]Lambda^{-1}
$$
answered Jan 7 at 3:06
OmnomnomnomOmnomnomnom
128k791183
$endgroup$
add a comment |
$begingroup$
Suppose that all $lambda_i$ are non-zero. One way to think of this action is to consider the fact that
$$
Lambda X = (Lambda X Lambda^{-1}) (Lambda LambdaLambda^{-1}) = [X]_{mathcal B} [Lambda]_{mathcal B}
$$
The transition $X mapsto Lambda X Lambda^{-1}$ can be thought of geometrically as a change in basis. That is, we reinterpret $X$ as a transformation relative to the basis $mathcal B = {lambda_1 e_1, dots, lambda_n e_n}$, where $e_i$ is the $i$th standard basis vector.
Multiplying from the right by $Lambda$, as before, amounts to scaling the columns.
Or, if you prefer: multiply first, then change basis, since
$$
Lambda X = Lambda [X Lambda]Lambda^{-1}
$$
answered Jan 7 at 3:06
OmnomnomnomOmnomnomnom
128k791183
$endgroup$
Suppose that all $lambda_i$ are non-zero. One way to think of this action is to consider the fact that
$$
Lambda X = (Lambda X Lambda^{-1}) (Lambda LambdaLambda^{-1}) = [X]_{mathcal B} [Lambda]_{mathcal B}
$$
The transition $X mapsto Lambda X Lambda^{-1}$ can be thought of geometrically as a change in basis. That is, we reinterpret $X$ as a transformation relative to the basis $mathcal B = {lambda_1 e_1, dots, lambda_n e_n}$, where $e_i$ is the $i$th standard basis vector.
Multiplying from the right by $Lambda$, as before, amounts to scaling the columns.
Or, if you prefer: multiply first, then change basis, since
$$
Lambda X = Lambda [X Lambda]Lambda^{-1}
$$
answered Jan 7 at 3:06
OmnomnomnomOmnomnomnom
128k791183
answered Jan 7 at 3:06
OmnomnomnomOmnomnomnom
128k791183
answered Jan 7 at 3:06
OmnomnomnomOmnomnomnom
128k791183
answered Jan 7 at 3:06
OmnomnomnomOmnomnomnom
128k791183
128k791183
add a comment |
add a comment |
$begingroup$
Notice that when you multiple $Lambda$ to X, the resultant matrix is $left[ lambda_1 x_1,,,,,lambda_2 x_2,,,ldots lambda_n x_nright]$. You multiply each column but the corresponding scalar.
The relationship between X$^prime$ and X is that it stretches (or contracts) each column vector.
answered Jan 7 at 2:29
Joel PereiraJoel Pereira
75919
$endgroup$
1
$begingroup$
Thank you for your reply. But isn't that a right multiplication of the diagonal matrix?
$endgroup$
– ArtificiallyIntelligence
Jan 7 at 2:32
add a comment |
$begingroup$
Notice that when you multiple $Lambda$ to X, the resultant matrix is $left[ lambda_1 x_1,,,,,lambda_2 x_2,,,ldots lambda_n x_nright]$. You multiply each column but the corresponding scalar.
The relationship between X$^prime$ and X is that it stretches (or contracts) each column vector.
answered Jan 7 at 2:29
Joel PereiraJoel Pereira
75919
$endgroup$
1
$begingroup$
Thank you for your reply. But isn't that a right multiplication of the diagonal matrix?
$endgroup$
– ArtificiallyIntelligence
Jan 7 at 2:32
add a comment |
$begingroup$
Notice that when you multiple $Lambda$ to X, the resultant matrix is $left[ lambda_1 x_1,,,,,lambda_2 x_2,,,ldots lambda_n x_nright]$. You multiply each column but the corresponding scalar.
The relationship between X$^prime$ and X is that it stretches (or contracts) each column vector.
answered Jan 7 at 2:29
Joel PereiraJoel Pereira
75919
$endgroup$
Notice that when you multiple $Lambda$ to X, the resultant matrix is $left[ lambda_1 x_1,,,,,lambda_2 x_2,,,ldots lambda_n x_nright]$. You multiply each column but the corresponding scalar.
The relationship between X$^prime$ and X is that it stretches (or contracts) each column vector.
answered Jan 7 at 2:29
Joel PereiraJoel Pereira
75919
answered Jan 7 at 2:29
Joel PereiraJoel Pereira
75919
answered Jan 7 at 2:29
Joel PereiraJoel Pereira
75919
answered Jan 7 at 2:29
Joel PereiraJoel Pereira
75919
75919
1
$begingroup$
Thank you for your reply. But isn't that a right multiplication of the diagonal matrix?
$endgroup$
– ArtificiallyIntelligence
Jan 7 at 2:32
add a comment |
1
$begingroup$
Thank you for your reply. But isn't that a right multiplication of the diagonal matrix?
$endgroup$
– ArtificiallyIntelligence
Jan 7 at 2:32
1
1
$begingroup$
Thank you for your reply. But isn't that a right multiplication of the diagonal matrix?
$endgroup$
– ArtificiallyIntelligence
Jan 7 at 2:32
$begingroup$
Thank you for your reply. But isn't that a right multiplication of the diagonal matrix?
$endgroup$
– ArtificiallyIntelligence
Jan 7 at 2:32
add a comment |
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