Change of basis | Chapter 13, Essence of linear algebra

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Recommends chaohuang · 2012-07 · Mathematics Stack Exchange · on “intuition for similar matrix”

My current understanding, not sure this helps: $B=PAP^{-1}$ means $BP=PA$, so if $A$ is a diagonal, $BP=PA$ is just the definition of eigenvetors, which is easy to understand. Because we know a matrix and its diagonal form are the same transformation under different basis, so to extrapolate to non-diagonal cases, we can define any square matrix of the form $B'P=PA'$ or $B'=PA'P^{-1}$ representing the same transformation under different basis, called similar matrix. Edit: $B=PAP^{-1}$ means $A=P^{-1}BP$, and think of $P$ as the change of basis matrix, then $P^{-1}BP$ means change of basis first, then do the linear transformation, then change the basis back, so $A$ and $B$ represent the same l

Recommends Brian61354270 · 2019-03 · Mathematics Stack Exchange · on “Linear maps and matrix representations”

No, any basis will do. These [LINK] videos [LINK] by Grant Sanderson may help you develop an intuition for why our choice in basis does not matter when describing a linear transformation. To summarize, each element $v_i$ of a vector $\mathbf v$ in $\mathbb R^n$ essential describes "how many" of the $i$ th basis vector to add up when constructing the vector $\mathbf v$ . In $\mathbb R^3$ , $$ \begin{bmatrix} a \\ b \\ c\end{bmatrix} = ae_1 + be_2 + ce_3 $$ where $\{e_1,e_2,e_3\}$ is any basis of $\mathbb R^3$ . Each column of a matrix $A \in \mathbb R^{n\times m}$ describes where the basis vectors of $\mathbb R^3$ "land" when moved through the transformation described by the matrix $A$ . The

Recommends TeWu · 2019-10 · Stack Overflow · on “How to convert a location on football pitch to coordinates on rectangle?”

If you want to learn more about linear algebra, definitely check out the "Essence of linear algebra" by 3Blue1Brown [LINK] on YouTube. This video [this video] is especially relevant in context of this question.

Recommends Joe Rakhimov · 2020-10 · Mathematics Stack Exchange · on “Matrix representation after change of basis”

I am trying to solve this task using Wolfram Notebook. I would like to know how new matrix is calculated (part b) after a change of basis. [LINK] Part a : I decided to show this part because part b is a continuation of part a Basis for y=-x: p := {{ 0, -1 },{-1, 0}} [LINK] 30 degrees rotation: RotationTransform[30*Degree, {0, 0}][{x, y}] [LINK] q is 30 degrees rotation as matrix: q:={{Sqrt[3]/2, -2^(-1)}, {1/2, Sqrt[3]/2}} [LINK] a:=p.q [LINK] Part b New basis: w:={{1,-1},{-1,2}} [LINK] I did not understand how exactly change of basis works, however, I tried to implement what I learned in this video [this video] : Inverse[w].p.w.a [LINK] This way seems to me not correct. Question: How can I calcu

Recommends user524813 · 2020-11 · Mathematics Stack Exchange · on “Find the change of coordinates matrix”

Getting the hang of changing basis can be really tricky so no worries about asking this question at all! Might I recommend watching the Youtuber "3blue1brown" and looking at his playlist essence of algebra. This video gives a really nice explanation for what is going on when we change basis: [this video] I will just give you answers for your problem with a little explanation: Firstly the matrix B that has columns equal to your basis vectors is : $$ B = \begin{pmatrix} 2 & 0 & 4 \\ -3 & 1 & 0 \\ 0 & 1 & -1 \\ \end{pmatrix} $$ Let us think about what B, as a map/function/transformation does. It takes some vector $v_B$ ( a vector writen in our B basis ) and gives us the vector $v_E$ (the vector writ

Recommends Darshan P. · 2021-11 · Mathematics Stack Exchange · on “Representing the cyclic differentiation pattern of $\frac{d^n}{dx^n}(\sin(x))$ using linear algebra.”

$$ \pmatrix{\frac{d^nS}{dx^n} \\ \frac {d^nC}{dx^n}} =\pmatrix{0 & -1 \\ 1 & \ 0 }^n\pmatrix{S \\ C }$$ How did I come to above formula? We can use the change of basis of vectors [this video] (thanks to $3$Blue$1$Brown YT channel [LINK] ) We know $\frac {d^nS}{dx^n} = S(x + \frac {n\pi}{2})$ and same for $\frac {d^nC}{dx^n} = C(x + \frac {n\pi}{2})$ Here we note the fact that every time we differentiate the result we get is the rotating axis with $90^0$ that is $\hat i \text { lands on } \hat j$ thus the same keeps happening every time we are differentiating. $\pmatrix {\cos\theta && \sin\theta\\-\sin\theta&&\cos\theta}_{\theta = \frac\pi2} = \pmatrix{0&&-1\\1&&0}$ Thus, for $n$ th time we have $

Recommends Darshan P. · 2021-11 · Mathematics Stack Exchange · on “What geometric transformation does this matrix apply on $\mathbb R^3$?”

You can watch the 3B1B-The change of basis [this video] and Linear Transformation [LINK] may this will help you get the intuition behind the scaling of $\hat i$ and $\hat j$ Also, I like to suggest curious people: You can watch all videos on *Essence Linear Algebra* [LINK]

Recommends Danziger · 2024-05 · Stack Overflow · on “Update canvas on mouse move”

You need to track the cursor position with mousemove [LINK] and then do a simple change of basis to translate screen coordinates to game grid coordinates: The cursor's position v_C = (x, y) is in the canonical basis, C . You have an alternate basis B (your 2.5D grid), and its alternate basis vectors in basis C are: columnVector_C = (cellHalfSizeLong, cellHalfSizeShort) rowsVector_C = (-cellHalfSizeLong, cellHalfSizeShort) You put them in columns on a matrix M_BC : const M_BC = [ [cellHalfSizeLong, -cellHalfSizeLong], [cellHalfSizeShort, cellHalfSizeShort], ]; This matrix helps you translate vectors from basis B to C Invert that matrix to get M_CB . This matrix helps you translate vectors fro

Cites Galen · 2020-04 · Stack Overflow · on “Question about applying PCA with one Component”

I see that you have a test set [LINK] in addition to a training set [LINK] , however this not the usual setup for PCA [LINK] . PCA [LINK] has multiple applications, but one of the main ones is dimensionality reduction [LINK] . Dimensionality reduction is about removing variables, and PCA serves this purpose by changing the basis [this video] of your data and ordering them by the amount (or relative amount) of the total variation that they linearly explain. Since this does not require test data, we can think of this as unsupervised machine learning [LINK] , although many would also prefer to call this feature engineering [LINK] as it is often used to preprocess data to improve the performance of m

Cites Martin · 2020-05 · Mathematics Stack Exchange · on “Question about 3blue1brown's vector basis video”

I've just watched this video [this video] , I was wondering, if we have basis \begin{bmatrix}2&-1\\1&1\end{bmatrix} When we take linear combinations of these basis' the columns correspond to both points in the x and y direction, i.e. the first column is the first vector that is 2 in the x direction and 1 in the y, but multiplying by vector \begin{bmatrix}x \\y \end{bmatrix} to give the linear system of equations \begin{bmatrix}2x-y\\x+y\end{bmatrix} the combination of the columns correspond to points in only one direction. Not sure if I've explained this well, any help would be appreciated, thanks.