# Row Echelon Form - Change of basis

Here is a row echelon form and original matrix:

My text says:

Our problem is to find bases for the column spaces of 𝑈 and 𝐴. Those spaces are different (just look at the matrices!) but their dimensions are the same.

How come the spaces the matrix spans have changed? No combination of the columns of U will ever give me anything in the third entry like A's columns so how are these two equivalent?

The first and third columns of 𝑈 are a basis for its column space. They are the columns with pivots. Every other column is a combination of those two. Furthermore, the same is true of the original 𝐴—even though its columns are different. The pivot columns of 𝐴 are a basis for its column space.

Has the basis of my matrix changed? Can anyone comment as to what happens to my original basis?

• Doing row operations changes the column space, in general.
– user403337
Jul 24, 2019 at 6:25
• But if those are operations that are supposed to preserve the original relation, then why does this happen? Jul 24, 2019 at 6:28
• In row operations, you are combining rows with different operations so you are preserving the row space (and also the null space) but not the column space. So for a basis for column space of $A$ look at the pivot columns in $U$, the corresponding columns in $A$ will form a basis for the column space of $A$. Jul 24, 2019 at 6:29
• It's an effect of doing row operations: relations between the rows can change. However, relations between corresponding columns are preserved, as are the row space and null space.
– user403337
Jul 24, 2019 at 6:34

They are equivalent because$$\begin{bmatrix}1 & 0 & 0 \\ -2 & 1 & 0 \\ 5 & -2 & 1\end{bmatrix}.A=U.$$And there is no such thing as the basis of a matrix. What happens is that the span of the columns of $$A$$ has the same dimension as the span of the columns of $$U$$. But it is quit easy to see that the span of the columns of $$U$$ is the space spanned by its first and third columns. So both column spaces have dimension $$2$$.
• Two matrices $A$ and $B$ are equivalent if there are invertible matrices $M$ and $N$ such that $A=M.B.N$. My computations show that your matrices $A$ and $U$ are equivalent. Jul 24, 2019 at 6:37
• No, I am not saying that they are similar. That would make no sense. And what I wrote is equivalent to$$\begin{bmatrix}1 & 0 & 0 \\ -2 & 1 & 0 \\ 5 & -2 & 1\end{bmatrix}.A.\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}=U.$$ Jul 24, 2019 at 6:41