Consider the following two types of multiplication (Please, Read the whole question):
$$ \begin{bmatrix} a & x & c \\ d & e & f \\ \end{bmatrix}. \begin{bmatrix} g\\ h\\i \end{bmatrix} $$


$$ \begin{bmatrix} a & x \\ d & e \\ f & g \end{bmatrix}. \begin{bmatrix} g\\ h \end{bmatrix} $$

Now, there's a rule in linear algebra that states:
"$T$ maps $\mathbb{R}^n$ onto $\mathbb{R}^m$ iff the columns of $A$ span $\mathbb{R}^m$"

Given the rule, the former matrix multiplication we see the fact that the multiplication results in the transformation of a three dimensional vector into two dimensions.

But, how is this possible, when the columns of the transformation matrix does not span $\mathbb{R}$ because the matrix is linearly dependent??

Similarly, the columns of the transformation matrix of the second multiplication is also not spanning $\mathbb{R}^3$ yet the vector of $\mathbb{R}^n$ with $n=2$ is being transformed to $\mathbb{R}^3$. How?

  • $\begingroup$ In your “rule”, replace “to” by “onto”. Then it’s correct. $\endgroup$ – Chris Godsil Feb 24 '18 at 17:05

In the first case your matrix is $A\in \mathbb{R}^{2\times 3}$ and the induced homomorphism is $$f_A:\mathbb{R}^3\to \mathbb{R}^2, \quad f_A(x)=Ax$$

The columns of $A$ are never linearly independent, only $2$ can be since they are in $\mathbb{R}^2$. If two columns are, a $2$ dimensional subspace of $\mathbb{R}^3$ is being mapped onto $\mathbb{R}^2$ and one dimension is "lost information". The vectors "lost" are being mappend into $(0,0)^T$, which is called the kernel of $f_A$ or $A$.

In the second case $B\in \mathbb{R}^{3\times 2}$ and $$f_B: \mathbb{R}^2\to \mathbb{R}^3,\quad f_B(x)=Bx$$ Here if the columns are linearly independent then the entire space $\mathbb{R}^2$ is mapped onto a 2 dimensional subspace inside $\mathbb{R}^3$. The image is a plane and the rest isn't being "touched" by $f_B $.

  • $\begingroup$ So, in both the cases, we draw a plane in R^3? $\endgroup$ – mathmaniage Feb 25 '18 at 7:25
  • $\begingroup$ How can we prove this algebraically? $\endgroup$ – mathmaniage Feb 25 '18 at 7:26
  • $\begingroup$ By knowing that the matrix is a representation of the underlying homomorphism with respect to a basis. If you treat the lin. Independent columns as the image of a basis then you can prove most of this I suppose. This will become obvious when you learn more linear algebra. $\endgroup$ – orange Feb 25 '18 at 7:36
  • $\begingroup$ So, i'd be okay leaving it for now? $\endgroup$ – mathmaniage Feb 25 '18 at 8:55

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