Exist a linear combination for full rank matrix? I met a problem this morning.

Suppose that there exists a full rank matrix $X = \begin{bmatrix} a &amp; b &amp; c \\ b &amp; c &amp; b\\ c &amp; d &amp; a  \end{bmatrix} \in \mathbb{R}^{3\times3}$. From the definition of linear independence we know that if the matrix is not a full rank matrix, a relationship would exist: $k_1X_1 + k_2X_2 = X_3$, where $X_1$, $X_2$ and $X_3$ are corresponding columns of $X$, and $k_1$, $k_2 \in \mathbb{R}\backslash0$.

If $k_1$ and $k_2$ exist, then:
$$\begin{bmatrix} X_1 &amp; X_2\end{bmatrix} \begin{bmatrix} k_1 \\ k_2\end{bmatrix}  = \begin{bmatrix} X_3 \end{bmatrix}$$
$$ \begin{bmatrix} k_1 \\ k_2\end{bmatrix}  = pinv (\begin{bmatrix} X_1 &amp; X_2\end{bmatrix}) \begin{bmatrix} X_3 \end{bmatrix}$$
$$ \begin{bmatrix} k_1 \\ k_2\end{bmatrix}  =  (\begin{bmatrix} X_1 \\X_2\end{bmatrix} \begin{bmatrix} X_1 &amp; X_2\end{bmatrix})^{-1}\begin{bmatrix} X_1\\ X_2\end{bmatrix} \begin{bmatrix} X_3 \end{bmatrix}$$

And I try some simulations in MATLAB, the $k_1$ and $k_2$ exist. But the first equation is not satisfied. 

So I am very confused. I must be wrong. But where? Could anyone tell me?
It seems you have solved a least square problem finding the projection of $X_3$ onto $span(X_1,X_2)$.

Refer to the related Is the pseudoinverse matrix the solution to the least squares problem?
Your characterization of linear dependence is incorrect. The columns of $X$ are linearly dependent iff there is a nontrivial linear combination $k_1X_1+k_2X_2+k_3X_3$ that vanishes, but this does not let you conclude that $k_1X_1+k_2X_2=X_3$. Consider, for example, $$X=\begin{bmatrix}1&amp;0&amp;0\\0&amp;0&amp;0\\0&amp;0&amp;1\end{bmatrix}.$$ The columns are linearly dependent (every set that contains the zero vector is), but the third column is obviously not a scalar multiple of the first.  

If the columns are linearly dependent, then you know that at least one of them is a linear combination of the other two, but you can’t a priori decide which one that is.