Inverse matrices, column space and null space | Chapter 7, Essence of linear algebra

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Strong rec 00100000 · 2024-11 · Mathematics Stack Exchange · on “How can I prove a zero-one matrix, that has all entries 1 except for the anti-diagonal, invertible?”

Any way of showing that a matrix is non-singular can show you that it is invertible. You can: Check that all of the column vectors are linearly independent Verify that the matrix has a nonzero determinant Put the matrix into row-echelon form and confirm that it is full rank among other methods. The third method can be done with only row operations. A really excellent video presenting the intuition behind this is this one [this video] in 3Blue1Brown's "Essence of Linear Algebra" series, which explains how you can interpret a singular linear transformation as one that compresses the vectors in a vector space into ones that all lie on a smaller vector space, so you lose some "information" about the

Recommends asddf · 2017-02 · Mathematics Stack Exchange · on “Why is a matrix invertible when its row-echelon form has no zero row?”

Hi i dont know if i can link vidoes but this one does give a pretty good intuitive understanding .https: [this video]

Recommends Ben Grossmann · 2017-04 · Mathematics Stack Exchange · on “Row Space and Column Space in Matrix”

I know you're not looking for online tutorials, but one more that you might find useful: I think the videos from 3blue1brown for chapter 3 [LINK] and chapter 6 [this video] will be useful. Each video is about 10 minutes long.

Recommends technazi · 2018-04 · Mathematics Stack Exchange · on “Word origin / meaning of 'kernel' in linear algebra”

As mentioned by @triple_sec it can be said that the kernel of a matrix consists of those vectors of the domain space that are mapped into the center (i.e., the origin) of the range space. These are those vectors that after undergoing a transformation, squished themselves into a zero vector because the determinant = 0. (That means the region squished into 0 volume); These vectors are also called free variables or non-pivot columns; N(A) = N(RREF(A)) The kernel of a transformation is equal to the kernel of RREF of the transformation. That is the vectors that get squished in an RREF or the original matrix are going to be equal. The pivots help us define the system of equations in reduced forms.

Recommends user90596 · 2019-01 · Mathematics Stack Exchange · on “I didn't understand 3blue1brown's video on inverse matrices, where a matrix has no inverse if its determinant is zero”

I was watching 3blue1brown's video on inverse matrices [this video] and I didn't understand what he said about the case where a matrix $A$ has no inverse when $\det(A)=0$ . Considering $Ax=v$ where $x$ and $v$ are vectors, he made two statements: If a transformation squishes all of plane onto a single line for a $2\times 2$ matrix then no transformation can unsquish that line back to a plane. (I didn't understand his explanation based on functions) It is still possible that solution exists even if inverse doesn't exist when $\det(A)=0$ if vector $v$ "lives" somewhere on that line. I don't understand why no such transformation exists in the first statement. I also don't understand what vector $v$

Recommends RajS · 2020-05 · Mathematics Stack Exchange · on “Nature of vectors transforming to point in 2D and 3D space”

While learning linear algebra from 3b1b series, I came across following sentence at this point [this video] : If a 2D tranformation squishes space onto a line, for example, there is a separate line in a different direction, full of vectors that get squished onto the origin. For example, in below diagram a transformation transforms all vectors on y-axis to the origin. [LINK] I have following doubts: Q1. Does all vectors that get squished to a line need to lie on a single line? For example, cant there be any transformation which squishes all green vectors to green dot on x-axis? Or all vectors should lie only on single line only if they get squished to the origin, that this restriction exists only

Recommends RajS · 2020-05 · Mathematics Stack Exchange · on “Understanding the column space”

I am learning linear algebra from 3b1b series [LINK] . At this point [this video] , he explains for what he is using variables $A, \vec{x}$ and $\vec{y}$ : [LINK] Then at here [this video] , he says: P1: [LINK] he further says following: Set of all possible outputs of matrix, whether it’s a line, a plane or a 3D space, whatever, is called the column space of your matrix. The column of matrix tell whether the basis vectors land: [LINK] P2: Span of those transformed basis vectors gives all possible outputs. [LINK] P3: In other words, the column space is the span of the columns of your matrix. (Note that all above $2\times 2$ matrices are different examples of $A$ .) Now the author of videos has taken un

Recommends Sage Stark · 2020-06 · Mathematics Stack Exchange · on “Geometric intuition behind inverse matrix restrictions”

We know that for an invertible matrix $A$ , $A^{-1} = \frac{\mathrm{adj}(A)}{\det{(A)}}$ . We can see algebraically that this is undefined when $\det(A)=0$ , but it is cryptic what this means geometrically. Is the following intuition valid? This is not meant to be rigorous. Define a matrix $A:\mathbb{R}^n \rightarrow \mathbb{R}^{n}$ (e.g. mapping 3D space onto the X-Y plane). Because this maps the $n$ -dimensional parallelepiped such that it does not occupy the $n$ -th dimension, it's $n$ -dimensional "volume" is $0$ . Hence, $\det(A)=0$ . Now, define another matrix $T:\mathbb{R}^{n-1}\rightarrow\mathbb{R}^{n-1}$ that maps the $(n-1)$ -dimensional parallelepiped such that it coincides with t

Recommends Bhartendu Kumar · 2021-05 · Mathematics Stack Exchange · on “How to Geometrically see the Null-space from the Geometric transformation meaning of Matrix?”

In this 3blue1brown [this video] video on null-spaces, he said and showed the collection of those points in the original vector space that after transformation by the matrix $A$ are moved to origin. I know the geometric transformation of a Matrix and how the original vector space is transformed by the Maytrix in to a new vector space. I know the Rank Calculation and solving Homogeneous equation using Jordan Reduction in Row echelon form. But How to geometrically find those points that will transform into origin, after applying the transformation? Like how the x-axis and y-axis are collapsing, from that information can we infer the nullspace, i.e. the points collapsing into origin? At least how th

Recommends Slim Shady · 2021-09 · Mathematics Stack Exchange · on “I don't understand why and how it (the hypothetical inverse) transforms a line into a "whole line of vectors". What does this even mean?”

I am watching the "Essence of Linear Algebra" Series by 3Blue1Brown. In this video [this video] , he explains how the determinants, inverse matrices and systems of linear equations are related. I don't understand a part starting on 6:43 and finishing on 7:07. In particular, he says that when a determinant is zero, the transformation transforms the 2-D space into a line. Now, when the transformation to a line is executed, there doesn't exist an inverse to this transformation. The reason for this is because there is not such function which would take a line and transform it into a plane. What he says exactly is the following: "You cannot un-squish a line to turn it into a plane. At least, that's no