Question about matrix equation and it's simplification
$begingroup$
For example, we had such an equation (where A is a matrix, and we need to find X, also where T is transpose) -
A = $(2 cdot X)^T$
So, we will have something like this? According to the sources which I found.
$$A = X^T cdot 2 cdot I^T$$
$$X^T = A cdot (2 cdot I^T)^{-1}$$
linear-algebra matrices determinant
edited Jan 17 at 18:13
Viktor Glombik
9081527
asked Jan 17 at 17:39
Aliaksei KlimovichAliaksei Klimovich
456
$endgroup$
add a comment |
$begingroup$
For example, we had such an equation (where A is a matrix, and we need to find X, also where T is transpose) -
A = $(2 cdot X)^T$
So, we will have something like this? According to the sources which I found.
$$A = X^T cdot 2 cdot I^T$$
$$X^T = A cdot (2 cdot I^T)^{-1}$$
linear-algebra matrices determinant
edited Jan 17 at 18:13
Viktor Glombik
9081527
asked Jan 17 at 17:39
Aliaksei KlimovichAliaksei Klimovich
456
$endgroup$
add a comment |
$begingroup$
For example, we had such an equation (where A is a matrix, and we need to find X, also where T is transpose) -
A = $(2 cdot X)^T$
So, we will have something like this? According to the sources which I found.
$$A = X^T cdot 2 cdot I^T$$
$$X^T = A cdot (2 cdot I^T)^{-1}$$
linear-algebra matrices determinant
edited Jan 17 at 18:13
Viktor Glombik
9081527
asked Jan 17 at 17:39
Aliaksei KlimovichAliaksei Klimovich
456
$endgroup$
For example, we had such an equation (where A is a matrix, and we need to find X, also where T is transpose) -
A = $(2 cdot X)^T$
So, we will have something like this? According to the sources which I found.
$$A = X^T cdot 2 cdot I^T$$
$$X^T = A cdot (2 cdot I^T)^{-1}$$
linear-algebra matrices determinant
linear-algebra matrices determinant
edited Jan 17 at 18:13
Viktor Glombik
9081527
asked Jan 17 at 17:39
Aliaksei KlimovichAliaksei Klimovich
456
edited Jan 17 at 18:13
Viktor Glombik
9081527
asked Jan 17 at 17:39
Aliaksei KlimovichAliaksei Klimovich
456
edited Jan 17 at 18:13
Viktor Glombik
9081527
edited Jan 17 at 18:13
Viktor Glombik
9081527
edited Jan 17 at 18:13
Viktor Glombik
9081527
9081527
asked Jan 17 at 17:39
Aliaksei KlimovichAliaksei Klimovich
456
asked Jan 17 at 17:39
Aliaksei KlimovichAliaksei Klimovich
456
asked Jan 17 at 17:39
Aliaksei KlimovichAliaksei Klimovich
456
456
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Scalars, such as 2 are not effected by transposition, since transposing only makes rows columns and the other way around, so multiplying all entries of the matrix with two before or after transposing doesn't make a difference:
$$
A = (2 X)^T = 2 X^T.
$$
More precisely one could write
$$
(2 X)^T = (2 I cdot X)^T
= X^T cdot (2 I)^T
= X^T cdot 2 (I)^T
= X^T cdot 2 (I)
= 2 X^T,
$$
but we still need the above demonstrated fact $(lambda A)^T = lambda A^T$.
Solving for $X$, we have
$$
A = 2 X^T
implies X^T = frac{1}{2} A
implies left(X^Tright)^T = left( frac{1}{2} A right)^T
implies X = frac{1}{2} A^T.
$$
Also, your way is correct, since your last result simplifies as follows.
$$
X^T = A cdot (2 cdot I^T)^{-1}
= A cdot 2^{-1} I^{-T}
= frac{1}{2} A,
$$
which is my second step from above.
answered Jan 17 at 18:00
Viktor GlombikViktor Glombik
9081527
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077275%2fquestion-about-matrix-equation-and-its-simplification%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Scalars, such as 2 are not effected by transposition, since transposing only makes rows columns and the other way around, so multiplying all entries of the matrix with two before or after transposing doesn't make a difference:
$$
A = (2 X)^T = 2 X^T.
$$
More precisely one could write
$$
(2 X)^T = (2 I cdot X)^T
= X^T cdot (2 I)^T
= X^T cdot 2 (I)^T
= X^T cdot 2 (I)
= 2 X^T,
$$
but we still need the above demonstrated fact $(lambda A)^T = lambda A^T$.
Solving for $X$, we have
$$
A = 2 X^T
implies X^T = frac{1}{2} A
implies left(X^Tright)^T = left( frac{1}{2} A right)^T
implies X = frac{1}{2} A^T.
$$
Also, your way is correct, since your last result simplifies as follows.
$$
X^T = A cdot (2 cdot I^T)^{-1}
= A cdot 2^{-1} I^{-T}
= frac{1}{2} A,
$$
which is my second step from above.
answered Jan 17 at 18:00
Viktor GlombikViktor Glombik
9081527
$endgroup$
add a comment |
$begingroup$
Scalars, such as 2 are not effected by transposition, since transposing only makes rows columns and the other way around, so multiplying all entries of the matrix with two before or after transposing doesn't make a difference:
$$
A = (2 X)^T = 2 X^T.
$$
More precisely one could write
$$
(2 X)^T = (2 I cdot X)^T
= X^T cdot (2 I)^T
= X^T cdot 2 (I)^T
= X^T cdot 2 (I)
= 2 X^T,
$$
but we still need the above demonstrated fact $(lambda A)^T = lambda A^T$.
Solving for $X$, we have
$$
A = 2 X^T
implies X^T = frac{1}{2} A
implies left(X^Tright)^T = left( frac{1}{2} A right)^T
implies X = frac{1}{2} A^T.
$$
Also, your way is correct, since your last result simplifies as follows.
$$
X^T = A cdot (2 cdot I^T)^{-1}
= A cdot 2^{-1} I^{-T}
= frac{1}{2} A,
$$
which is my second step from above.
answered Jan 17 at 18:00
Viktor GlombikViktor Glombik
9081527
$endgroup$
add a comment |
$begingroup$
Scalars, such as 2 are not effected by transposition, since transposing only makes rows columns and the other way around, so multiplying all entries of the matrix with two before or after transposing doesn't make a difference:
$$
A = (2 X)^T = 2 X^T.
$$
More precisely one could write
$$
(2 X)^T = (2 I cdot X)^T
= X^T cdot (2 I)^T
= X^T cdot 2 (I)^T
= X^T cdot 2 (I)
= 2 X^T,
$$
but we still need the above demonstrated fact $(lambda A)^T = lambda A^T$.
Solving for $X$, we have
$$
A = 2 X^T
implies X^T = frac{1}{2} A
implies left(X^Tright)^T = left( frac{1}{2} A right)^T
implies X = frac{1}{2} A^T.
$$
Also, your way is correct, since your last result simplifies as follows.
$$
X^T = A cdot (2 cdot I^T)^{-1}
= A cdot 2^{-1} I^{-T}
= frac{1}{2} A,
$$
which is my second step from above.
answered Jan 17 at 18:00
Viktor GlombikViktor Glombik
9081527
$endgroup$
Scalars, such as 2 are not effected by transposition, since transposing only makes rows columns and the other way around, so multiplying all entries of the matrix with two before or after transposing doesn't make a difference:
$$
A = (2 X)^T = 2 X^T.
$$
More precisely one could write
$$
(2 X)^T = (2 I cdot X)^T
= X^T cdot (2 I)^T
= X^T cdot 2 (I)^T
= X^T cdot 2 (I)
= 2 X^T,
$$
but we still need the above demonstrated fact $(lambda A)^T = lambda A^T$.
Solving for $X$, we have
$$
A = 2 X^T
implies X^T = frac{1}{2} A
implies left(X^Tright)^T = left( frac{1}{2} A right)^T
implies X = frac{1}{2} A^T.
$$
Also, your way is correct, since your last result simplifies as follows.
$$
X^T = A cdot (2 cdot I^T)^{-1}
= A cdot 2^{-1} I^{-T}
= frac{1}{2} A,
$$
which is my second step from above.
answered Jan 17 at 18:00
Viktor GlombikViktor Glombik
9081527
edited Jan 17 at 18:16
answered Jan 17 at 18:00
Viktor GlombikViktor Glombik
9081527
answered Jan 17 at 18:00
Viktor GlombikViktor Glombik
9081527
answered Jan 17 at 18:00
Viktor GlombikViktor Glombik
9081527
9081527
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077275%2fquestion-about-matrix-equation-and-its-simplification%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
