# Show that if A and B are an n n matrices such that AB =O (where O is the zero matrix) then A is not invertible or B is not invertible.

Question
Matrix transformations
Show that if A and B are an n n matrices such that AB =O (where O is the zero matrix) then A is not invertible or B is not invertible.

2020-12-04
If A is not invertible we are done. If A is invertible, we can multiply AB=0 by $$\displaystyle{A}^{{-{{1}}}}$$ from the left:
$$\displaystyle{A}^{{-{{1}}}}{\left({A}{B}\right)}={A}^{{-{{1}}}}{O}\to{B}={0}$$
So, B=0, meaning that B is not invertible.

### Relevant Questions

In the following question there are statements which are TRUE and statements which are FALSE.
Choose all the statements which are FALSE.
1. If the number of equations in a linear system exceeds the number of unknowns, then the system must be inconsistent - thus no solution.
2. If B has a column with zeros, then AB will also have a column with zeros, if this product is defined.
3. If AB + BA is defined, then A and B are square matrices of the same size/dimension/order.
4. Suppose A is an n x n matrix and assume A^2 = O, where O is the zero matrix. Then A = O.
5. If A and B are n x n matrices such that AB = I, then BA = I, where I is the identity matrix.
Zero Divisors If a and b are real or complex numbers such thal ab = O. then either a = 0 or b = 0. Does this property hold for matrices? That is, if A and Bare n x n matrices such that AB = 0. is il true lhat we must have A = 0 or B = 0? Prove lhe resull or find a counterexample.
a) Let A and B be symmetric matrices of the same size.
Prove that AB is symmetric if and only $$AB=BA.$$
b) Find symmetric $$2 \cdot 2$$
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Mark each of the following statement true or false: If A and B are matrices such that AB = O and $$A \neq O$$, then B = O.
Show that $$\displaystyle{C}^{\ast}={R}^{\ast}+\times{T},\text{where}\ {C}^{\ast}$$ is the multiplicative group of non-zero complex numbers, T is the group of complex numbers of modulus equal to 1, $$\displaystyle{R}^{\ast}+$$ is the multiplicative group of positive real numbers.
Let $$\displaystyle\le{f}{t}{\left\lbrace{v}_{{{1}}},\ {v}_{{{2}}},\dot{{s}},\ {v}_{{{n}}}{r}{i}{g}{h}{t}\right\rbrace}$$ be a basis for a vector space V. Prove that if a linear transformation $$\displaystyle{T}\ :\ {V}\rightarrow\ {V}$$ satisfies $$\displaystyle{T}{\left({v}_{{{1}}}\right)}={0}\ \text{for}\ {i}={1},\ {2},\dot{{s}},\ {n},$$ then T is the zero transformation.
Getting Started: To prove that T is the zero transformation, you need to show that $$\displaystyle{T}{\left({v}\right)}={0}$$ for every vector v in V.
(i) Let v be an arbitrary vector in V such that $$\displaystyle{v}={c}_{{{1}}}\ {v}_{{{1}}}\ +\ {c}_{{{2}}}\ {v}_{{{2}}}\ +\ \dot{{s}}\ +\ {c}_{{{n}}}\ {v}_{{{n}}}.$$
(ii) Use the definition and properties of linear transformations to rewrite $$\displaystyle{T}\ {\left({v}\right)}$$ as a linear combination of $$\displaystyle{T}\ {\left({v}_{{{1}}}\right)}$$.
(iii) Use the fact that $$\displaystyle{T}\ {\left({v}_{{i}}\right)}={0}$$ to conclude that $$\displaystyle{T}\ {\left({v}\right)}={0}$$, making T the zero tranformation.