Linear Algebra - Proofs involving Inverses

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Homework Help Overview

The discussion revolves around proofs related to the properties of invertible matrices in linear algebra, specifically focusing on conditions under which certain matrix products are non-invertible and the implications of singular matrices.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of a matrix being non-invertible, discussing the existence of non-zero vectors that satisfy the equation AB = 0. There is also inquiry into the relationship between the dimensions of matrices and their invertibility.

Discussion Status

Participants have engaged in clarifying the definitions of singular and non-singular matrices, with some suggesting methods to construct a non-zero matrix B. There is an ongoing exploration of the conditions under which the product of matrices remains non-invertible.

Contextual Notes

Some participants question the necessity of certain assumptions and the clarity of the reasoning presented, indicating a need for further explanation regarding the construction of matrix B and its properties.

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Two fairly simple proof problems. . . why aren't they simpler? :(

Homework Statement


Let A be an nxn matrix...
If A is not invertible then there exists an nxn matrix B such that AB = 0, B != 0. (not equal to)


Homework Equations


None really.


The Attempt at a Solution


Obviously, when A is the zero matrix, AB = 0.

If we call A the coefficient matrix in the system of equations Ax = 0, then x = x1B1 + x2B2 + ... + xnBn, where B = [B1|B2|...|Bn].

I can't seem to explain why that works. Is it obvious enough just to say that or is there a step of explanation I have left out?

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Homework Statement


If A is an m x n matrix, B is an n x m matrix and n < m, then AB is not invertible.


Homework Equations





The Attempt at a Solution


Obviously, A and B are not square and are therefore not invertible. Does that fact really matter? The product of invertible matrices is invertible, but is the product of non invertible matrices also non invertible?
 
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Well, what i would say is that: We know that an invertible matrix is nonsingular,moreover, a matrix is invertible if and only if it is non-singular. So, since here A is supposed to be non-invertible, it means that A is singular. What this means is that: There exists some non-zero vector, call it b such that Ab=0.
Extrapolating from this, we can argue that, there will be some non-zero vectors, call them

[tex]\arrow B_1, B_2,...B_n[/tex] such that [tex]A*B_i=0[/tex], for all i=1,2,...n.

So, if we built our matrix [tex]B=[B_1,B_2,...,B_n][/tex]

We have actuall proven that AB=0. Where as we can clearly see B is not the zero matrix.
 
How can I guarrantee that B is not zero?
 
You know there is a vector Ab=0 with b not equal to zero, as sutupidmath said. You also seem to know the column space of a matrix represents it's range. So build a matrix whose column space is only multiples of b.
 
Thanks everyone. Think I have it now.
 

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