Linear Algebra - Proofs involving Inverses

In summary, Two fairly simple proof problems are discussed. The first problem states that if a matrix A is not invertible, then there exists a matrix B such that AB=0. The second problem states that if A is an m x n matrix and B is an n x m matrix and n<m, then AB is not invertible. The solution to the first problem involves considering A as the coefficient matrix in the system of equations Ax=0 and using the fact that the product of non-invertible matrices is also non-invertible. The solution to the second problem involves understanding the concept of a singular matrix and using the column space to build a matrix B that proves AB=0.
  • #1
descendency
40
0
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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  • #2
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.
 
  • #3
How can I guarrantee that B is not zero?
 
  • #4
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.
 
  • #5
Thanks everyone. Think I have it now.
 

FAQ: Linear Algebra - Proofs involving Inverses

What is the definition of an inverse matrix in linear algebra?

An inverse matrix is a square matrix that when multiplied with the original matrix, results in the identity matrix. It is denoted by A-1 and is calculated by using the Gauss-Jordan elimination method or by using the adjugate matrix.

How do you prove that a matrix is invertible?

A matrix is invertible if its determinant is non-zero. This can be proven by using the determinant properties, such as the product rule and the fact that the determinant of the identity matrix is 1. If the determinant is non-zero, then the matrix has an inverse.

Can a non-square matrix have an inverse?

No, a non-square matrix cannot have an inverse. In order for a matrix to have an inverse, it must be a square matrix, meaning it has an equal number of rows and columns.

What is the relationship between the inverse of a matrix and its transpose?

The inverse of a matrix is equal to its transpose if and only if the matrix is orthogonal. In other words, the inverse of a matrix is equal to its transpose if and only if the columns of the matrix are orthonormal (orthogonal and unit length).

How do you use the inverse of a matrix to solve a system of linear equations?

To solve a system of linear equations using the inverse of a matrix, you must first set up the system in the form Ax = b, where A is the coefficient matrix, x is the variable matrix, and b is the constant matrix. Then, you can multiply both sides of the equation by the inverse of A, A-1, to solve for x.

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