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Problem is that i dont know if this is good solution to this problem or if i am missing something ? I would like to have some feedback.

Question:

We have matrix equation $Ax=b$ where $A$ is $n \times n$ matrix and equation $Ax=b$ has solution when $b \in \mathbb{R}^n$. Is matrix $A$ Nonsingular ? Provide explanation with answer.

Attempt to solve:

Matrix $A$ is nonsingular if Determinant is not zero.

$$Det(A)=|A| \neq 0$$

Meaning if $Det(A)\neq 0$ we have possibility to calculate inverse matrix for matrix $A$. There most be such $n \times n$-matrix $B$ that,

$$ AB=BA=I_{n} $$ $$ B=A^{-1} $$ Where $I_n$ is identity matrix. Meaning a singularmatrix that has value 1 in diagonal line and value of 0 elsewhere. Identity matrix is also ortogonal. $$\\$$ Equation Ax=b has solution if, $$ Ax=b $$ $$ x=A^{-1}b $$ In order to solve $Ax=b$ there has to be inverse matrix for $A$. Matrix $A$ has to be nonsingular if equation $Ax=b$ has every possible solution when $b \in \mathbb{R}^n$ $$\\$$ Any comment providing feedback would be much appreciated.

Thanks,

Tuki

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  • $\begingroup$ But note that $Ax=b$ can have a solution even if $A$ is singular, in some circumstances. For example, if $A=0$ and $b=0$, then every $x$ is a solution. $\endgroup$ – MPW Oct 6 '17 at 21:22
  • $\begingroup$ You've shown that if the matrix is non-singular, then there is solution for all $b$. But I think it asks you to show if there is a solution for all $b$, then the matrix is non-singular. $\endgroup$ – velut luna Oct 6 '17 at 21:25
  • $\begingroup$ What's the difference in between these two ? Don't they go both ways ? $\endgroup$ – Tuki Oct 6 '17 at 21:27
  • $\begingroup$ In case of it could be singular or nonsingular. it has solution in either case. It does mean that it could be either one right ? Answers provided on this post suggest that $DetA \neq 0$ is necessity ? Meaning that i am little confused right now. $\endgroup$ – Tuki Oct 6 '17 at 21:35
  • $\begingroup$ @Tuki There are many equivalent definitions of a matrix being non-singular. Which one are you using? $\endgroup$ – velut luna Oct 6 '17 at 21:37
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To show that if $Ax=b$ has solution for all $b$, then $A$ must be non-singular:

The equations $$Ax=e_i$$ has solution $x_i.$ Then $$A(x_1, x_2, \cdots, x_n)=(e_1, e_2, \cdots, e_n)=I$$ and hence $A$ has inverse and is non-singular.

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Since $Ax=b$ has solution for all $b \in \mathbb{R}^n$, the linear operator $L_A \colon \mathbb{R}^n \to \mathbb{R}^n$ given by $L_A(x):=Ax$ is surjective.

Therefore by the Rank-Nullity theorem it is must be also injective, so necessarily $\det A \neq 0$.

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  • $\begingroup$ Could you provide little insight of what is injective ? Also this is little hard to understand since i dont know what is Nullity-Rank but i'll try to look it up right now. $\endgroup$ – Tuki Oct 6 '17 at 21:32
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    $\begingroup$ A linear application is injective if and only if its kernel is $0$. For an application associated with a square matrice $A$, this amounts precisely to say that $A$ is non-singular. This is completely standard material, that you can find (together with the Rank-Nullity theorem) in every book on Linear Algebra. $\endgroup$ – Francesco Polizzi Oct 6 '17 at 21:36

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