If Ax = 0 has one solution, show m>=n

  • Thread starter WY
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In summary, the conversation discusses the problem of showing that m >= n given a matrix A with one solution for Ax=0. The expert explains that the Dimension Theorem for Matrices states that if a matrix has n columns, then the rank of A plus the nullity of A equals n. In this case, since the nullity of A is 0, the rank of A must be greater than or equal to n. They also suggest starting by looking at simple cases with matrices of 1x2 and 2x3. The expert also mentions that this problem can be approached from a linear equations perspective, where there are too many unknowns and not enough equations.
  • #1
WY
28
0
Hey

I'm not to sure on how to approach this problem - I think I have to use induction but I don't know where to start!
Question: Let A be an mxn matric. Suppose that Ax=0 has only one solution, namely x=0.
Show that m >= n


Thanks for the help in advance
 
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  • #2
Dimension theorem for Matrices says if matrix has n columns => rank(A)+nullity(A)=n
In your case nullity(A)=dimKer(A)=0
Is it possible rank(A) to be > min(m,n)?
 
  • #3
Thanks for replying :)
But I'm not to sure what you mean - i don't think i am very familiar with that theorem... could you elaborate a little? hehehe
 
  • #4
WY said:
Thanks for replying :)
But I'm not to sure what you mean - i don't think i am very familiar with that theorem... could you elaborate a little? hehehe

If you are unfamiliar with those formal properties and want to approach the problem from the basics, you could start by writing simple cases

[tex]
\left[ {\begin{array}{*{20}c}
a & b \\
\end{array}} \right]\left[ {\begin{array}{*{20}c}
{x_1 } \\
{x_2 } \\
\end{array}} \right] = 0
[/tex]


[tex]
\left[ {\begin{array}{*{20}c}
{\begin{array}{*{20}c}
a & b & c \\
\end{array}} \\
{\begin{array}{*{20}c}
d & e & f \\
\end{array}} \\
\end{array}} \right]\left[ {\begin{array}{*{20}c}
{x_1 } \\
{x_2 } \\
{x_3 } \\
\end{array}} \right] = 0
[/tex]

It's pretty easy to show that these equations can be satisfied with non-zero x vectors. From a linear equations perspective, it is matter of having too many unkowns and not enough equations.
 

1. How do you prove that a linear system Ax = 0 has only one solution?

In order to prove that a linear system Ax = 0 has only one solution, you must show that the columns of matrix A are linearly independent. This means that no column can be written as a linear combination of the other columns. If the columns are linearly independent, then the only solution to Ax = 0 is the trivial solution where all variables are equal to 0.

2. What does it mean for a matrix to have more rows than columns (m>=n)?

A matrix having more rows than columns (m>=n) means that the matrix is a tall matrix, with more equations than variables. This often occurs in systems of equations where there are more constraints than unknowns.

3. How does the number of solutions to a linear system relate to the number of rows and columns in the matrix?

The number of solutions to a linear system is determined by the number of rows and columns in the matrix. If there are more rows than columns (m>=n), the system will either have no solution or infinitely many solutions. If there are more columns than rows (m

4. What implications does having only one solution for Ax = 0 have on the matrix A?

If Ax = 0 has only one solution, it means that the matrix A has linearly independent columns. This has implications for the invertibility of the matrix, as well as the rank and nullity of the matrix.

5. Can a matrix have more rows than columns and still have only one solution for Ax = 0?

No, a matrix cannot have more rows than columns and still have only one solution for Ax = 0. This is because having more rows than columns (m>=n) means that the matrix is underdetermined and will either have no solution or infinitely many solutions.

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