Linear algebra: Find || || u || v ||
$begingroup$
Let u = (2, 4, 1) and v = (-3, 5, -1).
The question asks me to find || ||u||v|| but I don't know exactly what that means as I've never seen something written like this (other than guessing that it's a question about parallels). I've searched everywhere for a question similar to this but I couldn't find anything. Any help would be appreciated. Thanks!
linear-algebra
asked Jan 25 at 21:15
DwightDDwightD
42
$endgroup$
add a comment |
$begingroup$
Let u = (2, 4, 1) and v = (-3, 5, -1).
The question asks me to find || ||u||v|| but I don't know exactly what that means as I've never seen something written like this (other than guessing that it's a question about parallels). I've searched everywhere for a question similar to this but I couldn't find anything. Any help would be appreciated. Thanks!
linear-algebra
asked Jan 25 at 21:15
DwightDDwightD
42
$endgroup$
add a comment |
$begingroup$
Let u = (2, 4, 1) and v = (-3, 5, -1).
The question asks me to find || ||u||v|| but I don't know exactly what that means as I've never seen something written like this (other than guessing that it's a question about parallels). I've searched everywhere for a question similar to this but I couldn't find anything. Any help would be appreciated. Thanks!
linear-algebra
asked Jan 25 at 21:15
DwightDDwightD
42
$endgroup$
Let u = (2, 4, 1) and v = (-3, 5, -1).
The question asks me to find || ||u||v|| but I don't know exactly what that means as I've never seen something written like this (other than guessing that it's a question about parallels). I've searched everywhere for a question similar to this but I couldn't find anything. Any help would be appreciated. Thanks!
linear-algebra
linear-algebra
asked Jan 25 at 21:15
DwightDDwightD
42
asked Jan 25 at 21:15
DwightDDwightD
42
asked Jan 25 at 21:15
DwightDDwightD
42
asked Jan 25 at 21:15
DwightDDwightD
42
asked Jan 25 at 21:15
DwightDDwightD
42
42
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The notation of those double-bars refers to the norm on the vector space (usually the euclidean norm $||x||=sqrt{x_1^2+x_2^2+x_3^2}$). Knowing this, we can see how the expression in your question was constructed:
- Start with the vector $u$ and take its norm $||u||$.
$||u||$ is a real number - a scalar - so we can take any other vector $v$ and multiply it by that scalar to get $||u||v$.
$||u||v$ is a vector. Therefore, we can take its norm $||(||u||v)||$.
This is subtly different to the expression $||u||.||v||$ mentioned by Jagol95 in the comment. However, knowing the properties of norms it turns out that the two expressions are actually equal. I wouldn't be surprised if your teacher follows up by getting you to confirm this for the specific example (You should probably attempt this even if they don't).
answered Jan 25 at 21:25
ChessanatorChessanator
2,2061412
$endgroup$
add a comment |
$begingroup$
Hint: Probably $lVert mathbf{x}rVert$ signifies the (Euclidean) norm of the vector $mathbf{x}$, so
$lVert (x_1,x_2,x_3) rVert = sqrt{x_1^2+x_2^2+x_3^2}$.
You probably have a typo and mean $lVert mathbf{u}rVert lVert mathbf{v}rVert$, the product of $lVert mathbf{u}rVert$ and $lVert mathbf{v}rVert$.
Or, possibly you mean $BiglVert underbrace{lVert mathbf{u}rVert mathbf{v}}_textrm{scalar times vector} BigrVert$ (as @Chessanator suggests)? Your notation is difficult to understand without using MathJax.
answered Jan 25 at 21:23
MPWMPW
30.6k12157
$endgroup$
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%2f3087626%2flinear-algebra-find-u-v%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The notation of those double-bars refers to the norm on the vector space (usually the euclidean norm $||x||=sqrt{x_1^2+x_2^2+x_3^2}$). Knowing this, we can see how the expression in your question was constructed:
- Start with the vector $u$ and take its norm $||u||$.
$||u||$ is a real number - a scalar - so we can take any other vector $v$ and multiply it by that scalar to get $||u||v$.
$||u||v$ is a vector. Therefore, we can take its norm $||(||u||v)||$.
This is subtly different to the expression $||u||.||v||$ mentioned by Jagol95 in the comment. However, knowing the properties of norms it turns out that the two expressions are actually equal. I wouldn't be surprised if your teacher follows up by getting you to confirm this for the specific example (You should probably attempt this even if they don't).
answered Jan 25 at 21:25
ChessanatorChessanator
2,2061412
$endgroup$
add a comment |
$begingroup$
The notation of those double-bars refers to the norm on the vector space (usually the euclidean norm $||x||=sqrt{x_1^2+x_2^2+x_3^2}$). Knowing this, we can see how the expression in your question was constructed:
- Start with the vector $u$ and take its norm $||u||$.
$||u||$ is a real number - a scalar - so we can take any other vector $v$ and multiply it by that scalar to get $||u||v$.
$||u||v$ is a vector. Therefore, we can take its norm $||(||u||v)||$.
This is subtly different to the expression $||u||.||v||$ mentioned by Jagol95 in the comment. However, knowing the properties of norms it turns out that the two expressions are actually equal. I wouldn't be surprised if your teacher follows up by getting you to confirm this for the specific example (You should probably attempt this even if they don't).
answered Jan 25 at 21:25
ChessanatorChessanator
2,2061412
$endgroup$
add a comment |
$begingroup$
The notation of those double-bars refers to the norm on the vector space (usually the euclidean norm $||x||=sqrt{x_1^2+x_2^2+x_3^2}$). Knowing this, we can see how the expression in your question was constructed:
- Start with the vector $u$ and take its norm $||u||$.
$||u||$ is a real number - a scalar - so we can take any other vector $v$ and multiply it by that scalar to get $||u||v$.
$||u||v$ is a vector. Therefore, we can take its norm $||(||u||v)||$.
This is subtly different to the expression $||u||.||v||$ mentioned by Jagol95 in the comment. However, knowing the properties of norms it turns out that the two expressions are actually equal. I wouldn't be surprised if your teacher follows up by getting you to confirm this for the specific example (You should probably attempt this even if they don't).
answered Jan 25 at 21:25
ChessanatorChessanator
2,2061412
$endgroup$
The notation of those double-bars refers to the norm on the vector space (usually the euclidean norm $||x||=sqrt{x_1^2+x_2^2+x_3^2}$). Knowing this, we can see how the expression in your question was constructed:
- Start with the vector $u$ and take its norm $||u||$.
$||u||$ is a real number - a scalar - so we can take any other vector $v$ and multiply it by that scalar to get $||u||v$.
$||u||v$ is a vector. Therefore, we can take its norm $||(||u||v)||$.
This is subtly different to the expression $||u||.||v||$ mentioned by Jagol95 in the comment. However, knowing the properties of norms it turns out that the two expressions are actually equal. I wouldn't be surprised if your teacher follows up by getting you to confirm this for the specific example (You should probably attempt this even if they don't).
answered Jan 25 at 21:25
ChessanatorChessanator
2,2061412
answered Jan 25 at 21:25
ChessanatorChessanator
2,2061412
answered Jan 25 at 21:25
ChessanatorChessanator
2,2061412
answered Jan 25 at 21:25
ChessanatorChessanator
2,2061412
2,2061412
add a comment |
add a comment |
$begingroup$
Hint: Probably $lVert mathbf{x}rVert$ signifies the (Euclidean) norm of the vector $mathbf{x}$, so
$lVert (x_1,x_2,x_3) rVert = sqrt{x_1^2+x_2^2+x_3^2}$.
You probably have a typo and mean $lVert mathbf{u}rVert lVert mathbf{v}rVert$, the product of $lVert mathbf{u}rVert$ and $lVert mathbf{v}rVert$.
Or, possibly you mean $BiglVert underbrace{lVert mathbf{u}rVert mathbf{v}}_textrm{scalar times vector} BigrVert$ (as @Chessanator suggests)? Your notation is difficult to understand without using MathJax.
answered Jan 25 at 21:23
MPWMPW
30.6k12157
$endgroup$
add a comment |
$begingroup$
Hint: Probably $lVert mathbf{x}rVert$ signifies the (Euclidean) norm of the vector $mathbf{x}$, so
$lVert (x_1,x_2,x_3) rVert = sqrt{x_1^2+x_2^2+x_3^2}$.
You probably have a typo and mean $lVert mathbf{u}rVert lVert mathbf{v}rVert$, the product of $lVert mathbf{u}rVert$ and $lVert mathbf{v}rVert$.
Or, possibly you mean $BiglVert underbrace{lVert mathbf{u}rVert mathbf{v}}_textrm{scalar times vector} BigrVert$ (as @Chessanator suggests)? Your notation is difficult to understand without using MathJax.
answered Jan 25 at 21:23
MPWMPW
30.6k12157
$endgroup$
add a comment |
$begingroup$
Hint: Probably $lVert mathbf{x}rVert$ signifies the (Euclidean) norm of the vector $mathbf{x}$, so
$lVert (x_1,x_2,x_3) rVert = sqrt{x_1^2+x_2^2+x_3^2}$.
You probably have a typo and mean $lVert mathbf{u}rVert lVert mathbf{v}rVert$, the product of $lVert mathbf{u}rVert$ and $lVert mathbf{v}rVert$.
Or, possibly you mean $BiglVert underbrace{lVert mathbf{u}rVert mathbf{v}}_textrm{scalar times vector} BigrVert$ (as @Chessanator suggests)? Your notation is difficult to understand without using MathJax.
answered Jan 25 at 21:23
MPWMPW
30.6k12157
$endgroup$
Hint: Probably $lVert mathbf{x}rVert$ signifies the (Euclidean) norm of the vector $mathbf{x}$, so
$lVert (x_1,x_2,x_3) rVert = sqrt{x_1^2+x_2^2+x_3^2}$.
You probably have a typo and mean $lVert mathbf{u}rVert lVert mathbf{v}rVert$, the product of $lVert mathbf{u}rVert$ and $lVert mathbf{v}rVert$.
Or, possibly you mean $BiglVert underbrace{lVert mathbf{u}rVert mathbf{v}}_textrm{scalar times vector} BigrVert$ (as @Chessanator suggests)? Your notation is difficult to understand without using MathJax.
answered Jan 25 at 21:23
MPWMPW
30.6k12157
edited Jan 25 at 21:28
answered Jan 25 at 21:23
MPWMPW
30.6k12157
answered Jan 25 at 21:23
MPWMPW
30.6k12157
answered Jan 25 at 21:23
MPWMPW
30.6k12157
30.6k12157
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%2f3087626%2flinear-algebra-find-u-v%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
